$\text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. }$
-$$\lim _ { x \rightarrow0 } \frac { \tan 4 x } { x }$$
A) $\frac { 1 } { 4 }$
B) 4
C) 1
D) does not exist

Question

Quizplus · Accepted Answer

We can use the limit properties and algebraic manipulation to simplify the expression:

\begin{align*}
\lim_{xightarrow 0} \frac{	an 4x}{x} &= \lim_{xightarrow 0} \frac{\sin 4x}{x\cos 4x} \
&= 4\lim_{xightarrow 0} \frac{\sin 4x}{4x} \cdot \frac{1}{\cos 4x} \
&= 4\cdot1\cdot1 \
&= 4
\end{align*}

We used the identity $	an x = \frac{\sin x}{\cos x}$ and the fact that $\cos 0 = 1$ to simplify the expression. Since the limit exists and is finite, the answer is (B) 4.

Find the limit using lim⁡x=0sin⁡xx=1. \text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. } Find the limit using limx=0​xsinx​=1.

$\text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. }$