Inspired by Efren Medallo

According to the quotient rule in limits (see here ) , we have,

$\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{\displaystyle\lim_{x\to a}f(x)}{\displaystyle\lim_{x\to a}g(x)}\quad\textrm{if}\quad\lim_{x\to a}g(x)\neq 0$

In the problem here, it shows an incorrect use of this rule with $a=0$ and $g(x)=\sin x$ and $f(x)=x$ in LHS and $f(x)=0$ in RHS. Why is it an incorrect use? Since $\lim\limits_{x\to 0}\sin x=0$ which violates the "if" condition of the rule stated above.

Hence, we cannot just shove the limit value like that.

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Now, it's easy to show that both the limits exist and are finite, so I'm heading straight to evaluating them.

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First, let's evaluate the limit on the left side of the equation.

Approach 1:

Recall the Maclaurin series of $\sin x$ . We have,

$\sin x=\sum_{k=0}^\infty(-1)^k\frac{x^{2k+1}}{(2k+1)!}$

Hence, we have, when $x\to 0$ ,

$\sin x=x-\frac{x^3}{3!}+\mathcal{O}(x^5)\\\implies \frac{\sin x}{x}=1-\frac{x^2}{3!}+\mathcal{O}(x^4)=1\\ \implies \lim_{x\to 0}\frac{\sin x}{x}=1\implies \lim_{x\to 0}\frac{x}{\sin x}=\frac{1}{\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}}=\frac 11=1$

Approach 2:

This limit can also be proved to be $1$ using L'Hopital rule as follows:

$\lim_{x\to 0}\frac{x}{\sin x}=\lim_{x\to 0}\frac{1}{\cos x}=\frac{1}{\cos 0}=\frac 11=1$

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Now, moving onto the second limit, it's relatively simple. We have,

$\lim_{x\to 0}\frac{0}{\sin x}=\lim_{x\to 0}0=0$

Note that when $x\to 0$ , we have $\sin x\neq 0$ so there is no indeterminate case.

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Now, since we found out both the limits, we can see that the question states $1=0$ which is obviously false. Hence the answer. $_\square$

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Note: $~~\mathcal{O}(\cdot)$ denotes Big-O Notation.

0/k where k --> 0 but k > 0 will alway yield 0

x=0

sinx=sin0=0

(x/sinx)=(0/0)

so ,it is not right.....

By L'Hopital's rule, we can see that the LHS can be changed to $lim_{x\to0} \frac{1}{cos (x)}$ , which equals 1. On the other hand, the limit on the RHS is a constant 0. And there you have it!