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Calculus I / 105 practice questions

Choose the method.
Make the connection.

Draw the situation, choose a relationship, and work the problem before opening its visual solution. Search for a topic or mix the order to practice choosing the method.

Back to the course ↗

35 lesson question sets found · Page 1 of 6.

Foundations and limits

Functions describe a changing system.

Full visual lesson ↗
Calculate and explain

Rebuild the starting calculation

Calculate the position using input time 3 s, acceleration coefficient 1 m/s². State its sign and units.

Checked quantity: position · m. Follow the definition for finite-cutoff and numerical-model questions.

Hint

A function assigns one output to each allowed input. Track domain, range, and units before composing functions. For this cart, the coefficient multiplying time squared determines how quickly position grows.

Explore the visual solution
Motion illustration: position 9. Velocity direction shown where applicable.Position 9 · arrow: velocity direction
Time (s) versus Position (m). Position. Selected position: 9 m. A function assigns one output to each input. The graph is a relationship, not a picture of the track.00324648Position (m)Time (s)
Position

positions(3)=1(3)2=9 ms(3)=1(3)^2=9\,\mathrm m

A function assigns one output to each input. The graph is a relationship, not a picture of the track.

Assumptions, units, and limits

A cart starts at the origin with position at². The coefficient a is half the constant physical acceleration.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

A function assigns one output to each allowed input. Track domain, range, and units before composing functions. For this cart, the coefficient multiplying time squared determines how quickly position grows.

s(t)=at2,s:[0,8]→Rs(t)=at^2,\quad s:[0,8]\to\mathbb R
Calculate and explain

Apply the idea to a new scenario

At t=4t=4 s with a=0.5m/sa=0.5 \mathrm{m}/\mathrm{s}², find the cart position.

Checked quantity: position · m. Follow the definition for finite-cutoff and numerical-model questions.

Hint

A function assigns one output to each allowed input. Track domain, range, and units before composing functions. For this cart, the coefficient multiplying time squared determines how quickly position grows.

Explore the visual solution
Motion illustration: position 8. Velocity direction shown where applicable.Position 8 · arrow: velocity direction
Time (s) versus Position (m). Position. Selected position: 8 m. A function assigns one output to each input. The graph is a relationship, not a picture of the track.00164328Position (m)Time (s)
Position

positions(4)=0.5(4)2=8 ms(4)=0.5(4)^2=8\,\mathrm m

A function assigns one output to each input. The graph is a relationship, not a picture of the track.

Assumptions, units, and limits

A cart starts at the origin with position at². The coefficient a is half the constant physical acceleration.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

A function assigns one output to each allowed input. Track domain, range, and units before composing functions. For this cart, the coefficient multiplying time squared determines how quickly position grows.

s(t)=at2,s:[0,8]→Rs(t)=at^2,\quad s:[0,8]\to\mathbb R
Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual
Motion illustration: position 9. Velocity direction shown where applicable.Position 9 · arrow: velocity direction
Time (s) versus Position (m). Position. Selected position: 9 m. A function assigns one output to each input. The graph is a relationship, not a picture of the track.00324648Position (m)Time (s)
Position

positions(3)=1(3)2=9 ms(3)=1(3)^2=9\,\mathrm m

A function assigns one output to each input. The graph is a relationship, not a picture of the track.

Assumptions, units, and limits

A cart starts at the origin with position at². The coefficient a is half the constant physical acceleration.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. A function assigns one output to each allowed input. Track domain, range, and units before composing functions. For this cart, the coefficient multiplying time squared determines how quickly position grows.

Foundations and limits

Undo a measurement with an inverse.

Full visual lesson ↗
Calculate and explain

Rebuild the starting calculation

Calculate the recovered nonnegative input using measured output 9 m. State its sign and units.

Checked quantity: recovered nonnegative input · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

A function has an inverse on a domain where it is one-to-one. Swap the roles of input and output, solve for the old input, and carry the domain restriction with you. Function composition can verify the result.

Explore the visual solution
Read the modelf(x)=x2 (x≥0),f−1(y)=yf(x)=x^2\ (x\ge0),\quad f^{-1}(y)=\sqrt y

Restrict the square function to nonnegative inputs so it is one-to-one.

Input x versus Output x². Square, Inverse, Reflection line y=x. Selected recovered nonnegative input: 3 . The inverse recovers the input. Without the domain restriction, positive outputs would have two candidate inputs.0012.512.52525Output x²Input x
SquareInverse · dashedReflection line y=x · dashed

recovered nonnegative inputf−1(9)=9=3f^{-1}(9)=\sqrt{9}=3

The inverse recovers the input. Without the domain restriction, positive outputs would have two candidate inputs.

Assumptions, units, and limits

Restrict the square function to nonnegative inputs so it is one-to-one.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

A function has an inverse on a domain where it is one-to-one. Swap the roles of input and output, solve for the old input, and carry the domain restriction with you. Function composition can verify the result.

f(x)=x2 (x≥0),f−1(y)=yf(x)=x^2\ (x\ge0),\quad f^{-1}(y)=\sqrt y
Calculate and explain

Apply the idea to a new scenario

A sensor reading is 16 m on this nonnegative-input model. Recover the input.

Checked quantity: recovered nonnegative input · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

A function has an inverse on a domain where it is one-to-one. Swap the roles of input and output, solve for the old input, and carry the domain restriction with you. Function composition can verify the result.

Explore the visual solution
Read the modelf(x)=x2 (x≥0),f−1(y)=yf(x)=x^2\ (x\ge0),\quad f^{-1}(y)=\sqrt y

Restrict the square function to nonnegative inputs so it is one-to-one.

Input x versus Output x². Square, Inverse, Reflection line y=x. Selected recovered nonnegative input: 4 . The inverse recovers the input. Without the domain restriction, positive outputs would have two candidate inputs.0012.512.52525Output x²Input x
SquareInverse · dashedReflection line y=x · dashed

recovered nonnegative inputf−1(16)=16=4f^{-1}(16)=\sqrt{16}=4

The inverse recovers the input. Without the domain restriction, positive outputs would have two candidate inputs.

Assumptions, units, and limits

Restrict the square function to nonnegative inputs so it is one-to-one.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

A function has an inverse on a domain where it is one-to-one. Swap the roles of input and output, solve for the old input, and carry the domain restriction with you. Function composition can verify the result.

f(x)=x2 (x≥0),f−1(y)=yf(x)=x^2\ (x\ge0),\quad f^{-1}(y)=\sqrt y
Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual
Read the modelf(x)=x2 (x≥0),f−1(y)=yf(x)=x^2\ (x\ge0),\quad f^{-1}(y)=\sqrt y

Restrict the square function to nonnegative inputs so it is one-to-one.

Input x versus Output x². Square, Inverse, Reflection line y=x. Selected recovered nonnegative input: 3 . The inverse recovers the input. Without the domain restriction, positive outputs would have two candidate inputs.0012.512.52525Output x²Input x
SquareInverse · dashedReflection line y=x · dashed

recovered nonnegative inputf−1(9)=9=3f^{-1}(9)=\sqrt{9}=3

The inverse recovers the input. Without the domain restriction, positive outputs would have two candidate inputs.

Assumptions, units, and limits

Restrict the square function to nonnegative inputs so it is one-to-one.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. A function has an inverse on a domain where it is one-to-one. Swap the roles of input and output, solve for the old input, and carry the domain restriction with you. Function composition can verify the result.

Foundations and limits

Follow a value toward a missing point.

Full visual lesson ↗
Calculate and explain

Rebuild the starting calculation

Calculate the nearby function value (or limit at h=0h=0) using distance h from 2 0.1 . State its sign and units.

Checked quantity: nearby function value (or limit at h=0) · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Factor the numerator and cancel only for inputs where the canceled factor is nonzero. The simplified expression describes nearby values, so it can establish a limit without defining the original function at the missing point.

Explore the visual solution
Read the modelf(x)=x2−4x−2=x+2 (x≠2),lim⁡x→2f(x)=4f(x)=\frac{x^2-4}{x-2}=x+2\ (x\ne2),\quad\lim_{x\to2}f(x)=4

The original function is undefined at x=2.x=2. At h=0h=0 the readout shows its limit, not a newly defined function value.

Input x versus Function value. Extension with a hole at 2. Selected nearby function value (or limit at h=0): 4.1 . Approach from either side. Nearby values can settle even when the value at the target is missing.003264Function valueInput x
Extension with a hole at 2

nearby function value (or limit at h=0)f(2+h)=4+h=4.1,lim⁡h→0f(2+h)=4f(2+h)=4+h=4.1,\qquad\lim_{h\to0}f(2+h)=4

Approach from either side. Nearby values can settle even when the value at the target is missing.

Assumptions, units, and limits

The original function is undefined at x=2.x=2. At h=0h=0 the readout shows its limit, not a newly defined function value.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

Factor the numerator and cancel only for inputs where the canceled factor is nonzero. The simplified expression describes nearby values, so it can establish a limit without defining the original function at the missing point.

f(x)=x2−4x−2=x+2 (x≠2),lim⁡x→2f(x)=4f(x)=\frac{x^2-4}{x-2}=x+2\ (x\ne2),\quad\lim_{x\to2}f(x)=4
Calculate and explain

Apply the idea to a new scenario

Use h=−0.2h=-0.2 and find the nearby value at x=2+hx=2+h. Compare with the two-sided limit.

Checked quantity: nearby function value (or limit at h=0) · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Factor the numerator and cancel only for inputs where the canceled factor is nonzero. The simplified expression describes nearby values, so it can establish a limit without defining the original function at the missing point.

Explore the visual solution
Read the modelf(x)=x2−4x−2=x+2 (x≠2),lim⁡x→2f(x)=4f(x)=\frac{x^2-4}{x-2}=x+2\ (x\ne2),\quad\lim_{x\to2}f(x)=4

The original function is undefined at x=2.x=2. At h=0h=0 the readout shows its limit, not a newly defined function value.

Input x versus Function value. Extension with a hole at 2. Selected nearby function value (or limit at h=0): 3.8 . Approach from either side. Nearby values can settle even when the value at the target is missing.003264Function valueInput x
Extension with a hole at 2

nearby function value (or limit at h=0)f(2+h)=4+h=3.8,lim⁡h→0f(2+h)=4f(2+h)=4+h=3.8,\qquad\lim_{h\to0}f(2+h)=4

Approach from either side. Nearby values can settle even when the value at the target is missing.

Assumptions, units, and limits

The original function is undefined at x=2.x=2. At h=0h=0 the readout shows its limit, not a newly defined function value.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

Factor the numerator and cancel only for inputs where the canceled factor is nonzero. The simplified expression describes nearby values, so it can establish a limit without defining the original function at the missing point.

f(x)=x2−4x−2=x+2 (x≠2),lim⁡x→2f(x)=4f(x)=\frac{x^2-4}{x-2}=x+2\ (x\ne2),\quad\lim_{x\to2}f(x)=4
Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual
Read the modelf(x)=x2−4x−2=x+2 (x≠2),lim⁡x→2f(x)=4f(x)=\frac{x^2-4}{x-2}=x+2\ (x\ne2),\quad\lim_{x\to2}f(x)=4

The original function is undefined at x=2.x=2. At h=0h=0 the readout shows its limit, not a newly defined function value.

Input x versus Function value. Extension with a hole at 2. Selected nearby function value (or limit at h=0): 4.1 . Approach from either side. Nearby values can settle even when the value at the target is missing.003264Function valueInput x
Extension with a hole at 2

nearby function value (or limit at h=0)f(2+h)=4+h=4.1,lim⁡h→0f(2+h)=4f(2+h)=4+h=4.1,\qquad\lim_{h\to0}f(2+h)=4

Approach from either side. Nearby values can settle even when the value at the target is missing.

Assumptions, units, and limits

The original function is undefined at x=2.x=2. At h=0h=0 the readout shows its limit, not a newly defined function value.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. Factor the numerator and cancel only for inputs where the canceled factor is nonzero. The simplified expression describes nearby values, so it can establish a limit without defining the original function at the missing point.

Foundations and limits

A limit and a value must agree.

Full visual lesson ↗
Calculate and explain

Rebuild the starting calculation

Calculate the absolute mismatch with the limit using value assigned at x=2x=2 6 . State its sign and units.

Checked quantity: absolute mismatch with the limit · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

At a point, continuity requires a defined value, an existing two-sided limit, and equality between them. A removable discontinuity can be repaired by setting the missing or mismatched value equal to its limit.

Explore the visual solution
Read the modelf(x)={x+2x≠2cx=2,continuous at 2 iff c=4f(x)=\begin{cases}x+2&x\ne2\\c&x=2\end{cases},\quad\text{continuous at 2 iff }c=4

The left and right limits are both 4. Only the single assigned value changes.

Input x versus Function value. Nearby values. Selected absolute mismatch with the limit: 2 . Continuity requires a defined value, an existing two-sided limit, and agreement between them.003264Function valueInput x
Nearby values

absolute mismatch with the limit∣f(2)−lim⁡x→2f(x)∣=∣6−4∣=2|f(2)-\lim_{x\to2}f(x)|=|6-4|=2

Continuity requires a defined value, an existing two-sided limit, and agreement between them.

Assumptions, units, and limits

The left and right limits are both 4. Only the single assigned value changes.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

At a point, continuity requires a defined value, an existing two-sided limit, and equality between them. A removable discontinuity can be repaired by setting the missing or mismatched value equal to its limit.

f(x)={x+2x≠2cx=2,continuous at 2 iff c=4f(x)=\begin{cases}x+2&x\ne2\\c&x=2\end{cases},\quad\text{continuous at 2 iff }c=4
Calculate and explain

Apply the idea to a new scenario

Set the assigned value at x=2x=2 to 4. Find the mismatch with the limit.

Checked quantity: absolute mismatch with the limit · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

At a point, continuity requires a defined value, an existing two-sided limit, and equality between them. A removable discontinuity can be repaired by setting the missing or mismatched value equal to its limit.

Explore the visual solution
Read the modelf(x)={x+2x≠2cx=2,continuous at 2 iff c=4f(x)=\begin{cases}x+2&x\ne2\\c&x=2\end{cases},\quad\text{continuous at 2 iff }c=4

The left and right limits are both 4. Only the single assigned value changes.

Input x versus Function value. Nearby values. Selected absolute mismatch with the limit: 0 . Continuity requires a defined value, an existing two-sided limit, and agreement between them.003264Function valueInput x
Nearby values

absolute mismatch with the limit∣f(2)−lim⁡x→2f(x)∣=∣4−4∣=0|f(2)-\lim_{x\to2}f(x)|=|4-4|=0

Continuity requires a defined value, an existing two-sided limit, and agreement between them.

Assumptions, units, and limits

The left and right limits are both 4. Only the single assigned value changes.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

At a point, continuity requires a defined value, an existing two-sided limit, and equality between them. A removable discontinuity can be repaired by setting the missing or mismatched value equal to its limit.

f(x)={x+2x≠2cx=2,continuous at 2 iff c=4f(x)=\begin{cases}x+2&x\ne2\\c&x=2\end{cases},\quad\text{continuous at 2 iff }c=4
Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual
Read the modelf(x)={x+2x≠2cx=2,continuous at 2 iff c=4f(x)=\begin{cases}x+2&x\ne2\\c&x=2\end{cases},\quad\text{continuous at 2 iff }c=4

The left and right limits are both 4. Only the single assigned value changes.

Input x versus Function value. Nearby values. Selected absolute mismatch with the limit: 2 . Continuity requires a defined value, an existing two-sided limit, and agreement between them.003264Function valueInput x
Nearby values

absolute mismatch with the limit∣f(2)−lim⁡x→2f(x)∣=∣6−4∣=2|f(2)-\lim_{x\to2}f(x)|=|6-4|=2

Continuity requires a defined value, an existing two-sided limit, and agreement between them.

Assumptions, units, and limits

The left and right limits are both 4. Only the single assigned value changes.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. At a point, continuity requires a defined value, an existing two-sided limit, and equality between them. A removable discontinuity can be repaired by setting the missing or mismatched value equal to its limit.

Foundations and limits

A continuous response must cross intermediate levels.

Full visual lesson ↗
Calculate and explain

Rebuild the starting calculation

Calculate the crossing input c using target output l 2 . State its sign and units.

Checked quantity: crossing input c · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

The intermediate value theorem gives existence under continuity on a closed interval. It does not calculate the crossing. This strictly increasing cubic has a unique crossing; numerical bisection locates it by preserving opposite signs.

Explore the visual solution
Read the modelf(x)=x3+x,f(−1)=−2, f(2)=10,f(c)=L for some c∈(−1,2)f(x)=x^3+x,\quad f(-1)=-2,\ f(2)=10,\quad f(c)=L\text{ for some }c\in(-1,2)

The function is continuous on [−1,2]. Controls keep L strictly between the endpoint values; a bisection calculation locates the guaranteed crossing.

Input x versus x³+x. Continuous response, Target. Selected crossing input c: 1 . The intermediate value theorem guarantees a crossing but does not locate it or guarantee uniqueness in general. This example is strictly increasing, so the crossing is unique.-2-140.5102x³+xInput x
Continuous responseTarget · dashed

crossing input cc≈1,c3+c=2c\approx1,\quad c^3+c=2

The intermediate value theorem guarantees a crossing but does not locate it or guarantee uniqueness in general. This example is strictly increasing, so the crossing is unique.

Assumptions, units, and limits

The function is continuous on [−1,2]. Controls keep L strictly between the endpoint values; a bisection calculation locates the guaranteed crossing.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

The intermediate value theorem gives existence under continuity on a closed interval. It does not calculate the crossing. This strictly increasing cubic has a unique crossing; numerical bisection locates it by preserving opposite signs.

f(x)=x3+x,f(−1)=−2, f(2)=10,f(c)=L for some c∈(−1,2)f(x)=x^3+x,\quad f(-1)=-2,\ f(2)=10,\quad f(c)=L\text{ for some }c\in(-1,2)
Calculate and explain

Apply the idea to a new scenario

Set target L=4\mathrm{L}=4 and locate the crossing of x3+xx^3+x on [−1,2].

Checked quantity: crossing input c · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

The intermediate value theorem gives existence under continuity on a closed interval. It does not calculate the crossing. This strictly increasing cubic has a unique crossing; numerical bisection locates it by preserving opposite signs.

Explore the visual solution
Read the modelf(x)=x3+x,f(−1)=−2, f(2)=10,f(c)=L for some c∈(−1,2)f(x)=x^3+x,\quad f(-1)=-2,\ f(2)=10,\quad f(c)=L\text{ for some }c\in(-1,2)

The function is continuous on [−1,2]. Controls keep L strictly between the endpoint values; a bisection calculation locates the guaranteed crossing.

Input x versus x³+x. Continuous response, Target. Selected crossing input c: 1.3788 . The intermediate value theorem guarantees a crossing but does not locate it or guarantee uniqueness in general. This example is strictly increasing, so the crossing is unique.-2-140.5102x³+xInput x
Continuous responseTarget · dashed

crossing input cc≈1.3788,c3+c=4c\approx1.3788,\quad c^3+c=4

The intermediate value theorem guarantees a crossing but does not locate it or guarantee uniqueness in general. This example is strictly increasing, so the crossing is unique.

Assumptions, units, and limits

The function is continuous on [−1,2]. Controls keep L strictly between the endpoint values; a bisection calculation locates the guaranteed crossing.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

The intermediate value theorem gives existence under continuity on a closed interval. It does not calculate the crossing. This strictly increasing cubic has a unique crossing; numerical bisection locates it by preserving opposite signs.

f(x)=x3+x,f(−1)=−2, f(2)=10,f(c)=L for some c∈(−1,2)f(x)=x^3+x,\quad f(-1)=-2,\ f(2)=10,\quad f(c)=L\text{ for some }c\in(-1,2)
Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual
Read the modelf(x)=x3+x,f(−1)=−2, f(2)=10,f(c)=L for some c∈(−1,2)f(x)=x^3+x,\quad f(-1)=-2,\ f(2)=10,\quad f(c)=L\text{ for some }c\in(-1,2)

The function is continuous on [−1,2]. Controls keep L strictly between the endpoint values; a bisection calculation locates the guaranteed crossing.

Input x versus x³+x. Continuous response, Target. Selected crossing input c: 1 . The intermediate value theorem guarantees a crossing but does not locate it or guarantee uniqueness in general. This example is strictly increasing, so the crossing is unique.-2-140.5102x³+xInput x
Continuous responseTarget · dashed

crossing input cc≈1,c3+c=2c\approx1,\quad c^3+c=2

The intermediate value theorem guarantees a crossing but does not locate it or guarantee uniqueness in general. This example is strictly increasing, so the crossing is unique.

Assumptions, units, and limits

The function is continuous on [−1,2]. Controls keep L strictly between the endpoint values; a bisection calculation locates the guaranteed crossing.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. The intermediate value theorem gives existence under continuity on a closed interval. It does not calculate the crossing. This strictly increasing cubic has a unique crossing; numerical bisection locates it by preserving opposite signs.

Foundations and limits

A tiny angle connects arc length and height.

Full visual lesson ↗
Calculate and explain

Rebuild the starting calculation

Calculate the ratio value (limit at h=0h=0) using angle h 0.1 rad. State its sign and units.

Checked quantity: ratio value (limit at h=0) · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

On a unit circle, arc length equals the radian angle. Geometric comparison or standard squeeze bounds establish the sine ratio limit. Replacing an angle by a constant multiple changes the limiting coefficient.

Explore the visual solution
Read the modellim⁡h→0sin⁡hh=1\lim_{h\to0}\frac{\sin h}{h}=1

Angles are in radians. At zero the curve displays the continuous extension of the quotient.

Angle h (rad) versus sin(h)/h. Ratio (extension at 0). Selected ratio value (limit at h=0): 0.99833 . Radian measure is essential: arc length divided by radius makes the small-angle ratio approach one.0-10.5011sin(h)/hAngle h (rad)
Ratio (extension at 0)

ratio value (limit at h=0)sin⁡(0.1)0.1=0.998334\frac{\sin(0.1)}{0.1}=0.998334

Radian measure is essential: arc length divided by radius makes the small-angle ratio approach one.

Assumptions, units, and limits

Angles are in radians. At zero the curve displays the continuous extension of the quotient.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

On a unit circle, arc length equals the radian angle. Geometric comparison or standard squeeze bounds establish the sine ratio limit. Replacing an angle by a constant multiple changes the limiting coefficient.

lim⁡h→0sin⁡hh=1\lim_{h\to0}\frac{\sin h}{h}=1
Calculate and explain

Apply the idea to a new scenario

At h=0.25h=0.25 radians evaluate sin(h)/h and compare with one.

Checked quantity: ratio value (limit at h=0) · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

On a unit circle, arc length equals the radian angle. Geometric comparison or standard squeeze bounds establish the sine ratio limit. Replacing an angle by a constant multiple changes the limiting coefficient.

Explore the visual solution
Read the modellim⁡h→0sin⁡hh=1\lim_{h\to0}\frac{\sin h}{h}=1

Angles are in radians. At zero the curve displays the continuous extension of the quotient.

Angle h (rad) versus sin(h)/h. Ratio (extension at 0). Selected ratio value (limit at h=0): 0.98962 . Radian measure is essential: arc length divided by radius makes the small-angle ratio approach one.0-10.5011sin(h)/hAngle h (rad)
Ratio (extension at 0)

ratio value (limit at h=0)sin⁡(0.25)0.25=0.989616\frac{\sin(0.25)}{0.25}=0.989616

Radian measure is essential: arc length divided by radius makes the small-angle ratio approach one.

Assumptions, units, and limits

Angles are in radians. At zero the curve displays the continuous extension of the quotient.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

On a unit circle, arc length equals the radian angle. Geometric comparison or standard squeeze bounds establish the sine ratio limit. Replacing an angle by a constant multiple changes the limiting coefficient.

lim⁡h→0sin⁡hh=1\lim_{h\to0}\frac{\sin h}{h}=1
Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual
Read the modellim⁡h→0sin⁡hh=1\lim_{h\to0}\frac{\sin h}{h}=1

Angles are in radians. At zero the curve displays the continuous extension of the quotient.

Angle h (rad) versus sin(h)/h. Ratio (extension at 0). Selected ratio value (limit at h=0): 0.99833 . Radian measure is essential: arc length divided by radius makes the small-angle ratio approach one.0-10.5011sin(h)/hAngle h (rad)
Ratio (extension at 0)

ratio value (limit at h=0)sin⁡(0.1)0.1=0.998334\frac{\sin(0.1)}{0.1}=0.998334

Radian measure is essential: arc length divided by radius makes the small-angle ratio approach one.

Assumptions, units, and limits

Angles are in radians. At zero the curve displays the continuous extension of the quotient.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. On a unit circle, arc length equals the radian angle. Geometric comparison or standard squeeze bounds establish the sine ratio limit. Replacing an angle by a constant multiple changes the limiting coefficient.