# How to get $\lim_{x\to 0}\frac{x}{\sin x} =1$ from $\lim_{x\to 0}\frac{\sin x}{x} =1$?

We know the identity that

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

However in many solved examples that I was going through , I came across the identity

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

Although it was never formally mentioned anywhere in the text . How does the previous identity imply this?

Does it mean that as $x$ becomes very small the value of $\sin x$ is approximately equal to the value of $x$ so the value of both of $\frac{x}{\sin x}$ and $\frac{\sin x}{x}$ tends to $1$?

• $g(x)=\frac{1}{x}$ is continuous in a neighbourhood of $x=1$, hence $\lim_{x\to 0}f(x)=1$ is equivalent to $\lim_{x\to 0}\frac{1}{f(x)}=1$. Commented Mar 2, 2018 at 15:27

By the Algebraic Limit Theorem $$\lim_{x \rightarrow0}1=1$$ $$\lim_{x \rightarrow0}\dfrac{\sin{x}}{x}=1$$

$$\implies \lim_{x\rightarrow 0}\dfrac{1}{\dfrac{\sin{x}}{x}}=\lim_{x \rightarrow0}\dfrac{x}{\sin{x}}=\frac{1}{1}=1$$

This can be used since $\lim_{x\rightarrow 0}\dfrac{\sin{x}}{x}$ is known and is not $0$

• Alright ! I understood it. Thank you ! Commented Mar 2, 2018 at 11:37

Simply note that

$$\frac{x}{\sin x}=\frac1{\frac{\sin x}{x}}\to \frac11=1$$

You could have tried L'Hospital here instead. $$\lim_{x\rightarrow 0}\frac {x}{\sin x}=\lim_{x\rightarrow 0}\frac {1}{\cos x}=1$$

Explanation

Consider $f(x)=x$ and $g(x)=\sin x$ and it is easily identifiable that $\frac {f(x)}{g(x)}\rightarrow \frac {0}{0}$ form when $x\rightarrow 0$.

Hence $$\lim_{x\rightarrow 0} \frac {f(x)}{g(x)}=\lim_{x\rightarrow 0} \frac {f'(x)}{g'(x)}=\lim_{x\rightarrow 0}\frac {1}{\cos x}=1$$

• @Aditi I have posted the answer which uses L'Hospital rule because since you are learning limits and you have reached at indeterminate forms so I thought you might have learned about it. If not I apologise for answer but you can surely refer to other answers if you didn't learn about L'Hospital Commented Mar 2, 2018 at 11:43
• Yes you are correct. I just started learning limits yesterday and I know the L’Hospital Rule. But I can’t say for sure that I know how to use it very well. I have yet to learn it completely. Anyways thank you ! Commented Mar 2, 2018 at 11:48

You have the limit identities such as

$$\lim (f(x) + g(x)) = \lim f(x) + \lim g(x)$$

$$\lim f(x) g(x) = (\lim f(x) )( \lim g(x))$$

and

$$\lim \frac{f(x)}{g(x))} = \frac{\lim f(x)}{\lim g(x)}$$

etc., which hold as long as the individual limits exist (and in the last case, we need $\lim g(x) \neq 0.$)

So you have the tools to compute:

$$\lim \frac{x}{\sin x} = \lim \frac{1}{\frac{\sin x}{x} } =\frac{\lim 1}{\lim \frac{\sin x}{x}} = \frac{1}{1}.$$

• @FedericoPoloni Thanks. FIxed. Commented Mar 2, 2018 at 13:51

The intuitive meaning of

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

is that for $x$ small, $\frac{\sin x}{x} \approx 1.$

Equivalently, $\sin x \approx x$, which of course implies $x \approx \sin x$ and hence $\frac{x}{\sin x} \approx 1$, which is the intuitive meaning of

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

This isn't a proof (for a proof, see any of the other answers) but it shows that the equivalence of the two limits should be intuitively obvious rather than surprising.

Interestingly, most calculus books first prove the second limit (the one with $\sin x$ in the denominator) and then take reciprocals to get the first limit. The reason for turning the limit upside down like that it is that form which arises when using the limit definition of derivatives to get the derivatives of sine and cosine.

• Exactly what I was thinking . Thank you ! Commented Mar 2, 2018 at 16:56

You can also use Taylor expansion: $$\sin x=x \ + \ o(x)$$ Then $$\lim_{x\to 0}\frac{\sin x}{x}=\lim_{x\to 0}\frac{x \ + \ o(x)}{x}=\lim_{x\to 0}\frac{x}{x}=\lim_{x\to 0}\frac{x}{\sin x}=\lim_{x\to 0}\frac{x}{x \ + \ o(x)}=\lim_{x\to 0}\frac{x}{x}=1$$

• In regard to one of your other edits, note this difference: $$1+\frac{1}{2+\frac{2}{3+\frac{3}{4+\frac{4}{5+\frac{5}{...}}}}}$$ $$1+\cfrac{1}{2+\cfrac{2}{3+\cfrac{3}{4+\cfrac{4}{5+\cfrac{5}{...}}}}}$$ $$1+\cfrac{1}{2+\cfrac{2}{3+\cfrac{3}{4+\cfrac{4}{5+\cfrac{5}{\ddots}}}}}$$ The first form above uses \frac; the second and third use \cfrac; the third uses \ddots $\qquad$ Commented Mar 19, 2018 at 16:31