Limits - Math

Card 0 of 8

Question

Calculate $\lim_{x \rightarrow 0}\frac{\sin 4 x^{2} }{x^{2}}$

Answer

You can substitute $u = 4x^{2}$ to write this as:

$\lim_{x \rightarrow 0}\frac{\sin 4 x^{2} }{x^{2}}$

$= \lim_{x \rightarrow 0}\left (4 \cdot \frac{\sin 4 x^{2} }{4x^{2}} \right )$

$=4 \lim_{x \rightarrow 0} \frac{\sin 4 x^{2} }{4x^{2}}$

$=4 \lim_{u \rightarrow 0} \frac{\sin u }{u}$

Note that as $x \rightarrow 0$ , $u = 4x^{2} \rightarrow 0$

$\lim_{u \rightarrow 0} \frac{\sin u }{u}$ , since the fraction becomes indeterminate, we need to take the derivative of both the top and bottom of the fraction.

$\lim_{u \rightarrow 0} \frac{\sin u }{u}=\frac{(\sin u)'}{u'}=\frac{\cos u}{1}=\cos u$

$4\lim_{u\rightarrow 0}\cos u=4\cdot \cos(0)=4\cdot 1$

$4 \lim_{u \rightarrow 0} \frac{\sin u }{u} = 4 \cdot 1 = 4$ , which is the correct choice.

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Question

Calculate $\lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{x^{2}-1}$ .

Answer

Substitute $u = 2x^{2} - 2$ to rewrite this limit in terms of u instead of x. Multiply the top and bottom of the fraction by 2 in order to make this substitution:

(Note that as $x \rightarrow1$ , $u \rightarrow 0$ .)

$\lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{x^{2}-1}$

$=2 \cdot \lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{2\left (x^{2}-1 \right )}$

$= 2 \cdot \lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{2x^{2}-2 \right )}$

$= 2 \cdot \lim_{u \rightarrow 0}\frac{\sin u } {u}$

$\lim_{u \rightarrow 0} \frac{\sin u }{u} = 1$ , so

$2 \cdot \lim_{u \rightarrow 0}\frac{\sin u } {u} = 2\cdot1 = 2$ , which is therefore the correct answer choice.

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Question

Calculate $\lim_{x \rightarrow 0}\frac{\sin 4 x^{2} }{x^{2}}$

Answer

You can substitute $u = 4x^{2}$ to write this as:

$\lim_{x \rightarrow 0}\frac{\sin 4 x^{2} }{x^{2}}$

$= \lim_{x \rightarrow 0}\left (4 \cdot \frac{\sin 4 x^{2} }{4x^{2}} \right )$

$=4 \lim_{x \rightarrow 0} \frac{\sin 4 x^{2} }{4x^{2}}$

$=4 \lim_{u \rightarrow 0} \frac{\sin u }{u}$

Note that as $x \rightarrow 0$ , $u = 4x^{2} \rightarrow 0$

$\lim_{u \rightarrow 0} \frac{\sin u }{u}$ , since the fraction becomes indeterminate, we need to take the derivative of both the top and bottom of the fraction.

$\lim_{u \rightarrow 0} \frac{\sin u }{u}=\frac{(\sin u)'}{u'}=\frac{\cos u}{1}=\cos u$

$4\lim_{u\rightarrow 0}\cos u=4\cdot \cos(0)=4\cdot 1$

$4 \lim_{u \rightarrow 0} \frac{\sin u }{u} = 4 \cdot 1 = 4$ , which is the correct choice.

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Question

Calculate $\lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{x^{2}-1}$ .

Answer

Substitute $u = 2x^{2} - 2$ to rewrite this limit in terms of u instead of x. Multiply the top and bottom of the fraction by 2 in order to make this substitution:

(Note that as $x \rightarrow1$ , $u \rightarrow 0$ .)

$\lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{x^{2}-1}$

$=2 \cdot \lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{2\left (x^{2}-1 \right )}$

$= 2 \cdot \lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{2x^{2}-2 \right )}$

$= 2 \cdot \lim_{u \rightarrow 0}\frac{\sin u } {u}$

$\lim_{u \rightarrow 0} \frac{\sin u }{u} = 1$ , so

$2 \cdot \lim_{u \rightarrow 0}\frac{\sin u } {u} = 2\cdot1 = 2$ , which is therefore the correct answer choice.

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Question

Calculate $\lim_{x \rightarrow 0}\frac{\sin 4 x^{2} }{x^{2}}$

Answer

You can substitute $u = 4x^{2}$ to write this as:

$\lim_{x \rightarrow 0}\frac{\sin 4 x^{2} }{x^{2}}$

$= \lim_{x \rightarrow 0}\left (4 \cdot \frac{\sin 4 x^{2} }{4x^{2}} \right )$

$=4 \lim_{x \rightarrow 0} \frac{\sin 4 x^{2} }{4x^{2}}$

$=4 \lim_{u \rightarrow 0} \frac{\sin u }{u}$

Note that as $x \rightarrow 0$ , $u = 4x^{2} \rightarrow 0$

$\lim_{u \rightarrow 0} \frac{\sin u }{u}$ , since the fraction becomes indeterminate, we need to take the derivative of both the top and bottom of the fraction.

$\lim_{u \rightarrow 0} \frac{\sin u }{u}=\frac{(\sin u)'}{u'}=\frac{\cos u}{1}=\cos u$

$4\lim_{u\rightarrow 0}\cos u=4\cdot \cos(0)=4\cdot 1$

$4 \lim_{u \rightarrow 0} \frac{\sin u }{u} = 4 \cdot 1 = 4$ , which is the correct choice.

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Question

Calculate $\lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{x^{2}-1}$ .

Answer

Substitute $u = 2x^{2} - 2$ to rewrite this limit in terms of u instead of x. Multiply the top and bottom of the fraction by 2 in order to make this substitution:

(Note that as $x \rightarrow1$ , $u \rightarrow 0$ .)

$\lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{x^{2}-1}$

$=2 \cdot \lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{2\left (x^{2}-1 \right )}$

$= 2 \cdot \lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{2x^{2}-2 \right )}$

$= 2 \cdot \lim_{u \rightarrow 0}\frac{\sin u } {u}$

$\lim_{u \rightarrow 0} \frac{\sin u }{u} = 1$ , so

$2 \cdot \lim_{u \rightarrow 0}\frac{\sin u } {u} = 2\cdot1 = 2$ , which is therefore the correct answer choice.

Compare your answer with the correct one above

Question

Calculate $\lim_{x \rightarrow 0}\frac{\sin 4 x^{2} }{x^{2}}$

Answer

You can substitute $u = 4x^{2}$ to write this as:

$\lim_{x \rightarrow 0}\frac{\sin 4 x^{2} }{x^{2}}$

$= \lim_{x \rightarrow 0}\left (4 \cdot \frac{\sin 4 x^{2} }{4x^{2}} \right )$

$=4 \lim_{x \rightarrow 0} \frac{\sin 4 x^{2} }{4x^{2}}$

$=4 \lim_{u \rightarrow 0} \frac{\sin u }{u}$

Note that as $x \rightarrow 0$ , $u = 4x^{2} \rightarrow 0$

$\lim_{u \rightarrow 0} \frac{\sin u }{u}$ , since the fraction becomes indeterminate, we need to take the derivative of both the top and bottom of the fraction.

$\lim_{u \rightarrow 0} \frac{\sin u }{u}=\frac{(\sin u)'}{u'}=\frac{\cos u}{1}=\cos u$

$4\lim_{u\rightarrow 0}\cos u=4\cdot \cos(0)=4\cdot 1$

$4 \lim_{u \rightarrow 0} \frac{\sin u }{u} = 4 \cdot 1 = 4$ , which is the correct choice.

Compare your answer with the correct one above

Question

Calculate $\lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{x^{2}-1}$ .

Answer

Substitute $u = 2x^{2} - 2$ to rewrite this limit in terms of u instead of x. Multiply the top and bottom of the fraction by 2 in order to make this substitution:

(Note that as $x \rightarrow1$ , $u \rightarrow 0$ .)

$\lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{x^{2}-1}$

$=2 \cdot \lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{2\left (x^{2}-1 \right )}$

$= 2 \cdot \lim_{x \rightarrow 1}\frac{\sin \left (2x^{2}-2 \right ) }{2x^{2}-2 \right )}$

$= 2 \cdot \lim_{u \rightarrow 0}\frac{\sin u } {u}$

$\lim_{u \rightarrow 0} \frac{\sin u }{u} = 1$ , so

$2 \cdot \lim_{u \rightarrow 0}\frac{\sin u } {u} = 2\cdot1 = 2$ , which is therefore the correct answer choice.

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