Limits - Math

Card 0 of 8

Calculate

Tap to see back →

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

$= $\lim_{x \rightarrow 0}$\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$

, since the fraction becomes indeterminate, we need to take the derivative of both the top and bottom of the fraction.

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

Calculate .

Tap to see back →

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 $x \rightarrow1$ , $u \rightarrow 0$ .)

$=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}$

, so

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

Calculate

Tap to see back →

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

$= $\lim_{x \rightarrow 0}$\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$

, since the fraction becomes indeterminate, we need to take the derivative of both the top and bottom of the fraction.

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

Calculate .

Tap to see back →

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 $x \rightarrow1$ , $u \rightarrow 0$ .)

$=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}$

, so

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

Calculate

Tap to see back →

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

$= $\lim_{x \rightarrow 0}$\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$

, since the fraction becomes indeterminate, we need to take the derivative of both the top and bottom of the fraction.

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

Calculate .

Tap to see back →

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 $x \rightarrow1$ , $u \rightarrow 0$ .)

$=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}$

, so

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

Calculate

Tap to see back →

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

$= $\lim_{x \rightarrow 0}$\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$

, since the fraction becomes indeterminate, we need to take the derivative of both the top and bottom of the fraction.

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

Calculate .

Tap to see back →

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 $x \rightarrow1$ , $u \rightarrow 0$ .)

$=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}$

, so

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