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

Question

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 $1$?

Answer 1

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.

Answer 2

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

Answer 3

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 $0$

Answer 4

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:

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

Answer 5

Simply note that

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

Answer 6

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