3

I was solving a problem today and at one point I had to evaluate the $\lim_{x \to 0}{\frac {x}{\sin x}}$. I know I could easily do this with L'hôspital's but I haven't learned that yet.

So what I did was try to use the squeeze theorem with the two bounding functions of $ f (x)=x^2 + 1 $ and $ h (x)=-x^2 +1 $, and it worked (I think?) giving me the answer of 1. However, I was only able to choose those two functions after googling the graph of $\lim_{x \to 0}{\frac {x}{\sin x}}$, so that kind of ruins the purpose. Is there a way I could have visualized this or chosen other, more fitting functions?

Or maybe there is another way to solve this limit?

Martin Sleziak
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Howcan
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2 Answers2

3

The hint given by OC-Sansoo is not enough, which only shows that $\lim_{x\to 0^+}\frac{\sin x}{x}=0$.

What you should do is to prove that $\cos x < \dfrac{\sin x}{x} < 1$ for $x \in (-\pi/2,0)$.

Since $x \in (-\pi/2,0)$, $-x \in (0, \pi/2)$, hence, $\cos (-x) < \dfrac{\sin (-x)}{-x} < 1$, and hence $\cos x < \dfrac{\sin x}{x} < 1$.

Another way:

$$\lim_{x\to 0}\frac{\sin x}{x}=\lim_{x\to 0}\frac{\sin x-\sin 0}{x-0}=\sin'(x)\big|_{x=0}=\cos x|_{x=0}=\cos 0=1$$

Paul
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2

Hint: Prove that: $\cos x < \dfrac{\sin x}{x} < 1$ for $x \in (0,\pi/2)$

DeepSea
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