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I'm getting stuck on the following exercise where I have to find the limit as x approaches zero, for this cosine function:

$$\lim_{x \to 0}\cos\left(\frac{\pi\sin^2(x)}{x^2}\right)$$

The graph shows that there should be a limit of $-1$ at $0$, but I can't find a nice trigonometric identity that allows me to rewrite this such that the $x^2$ in the denominator disappears.

Any indication on how to solve this?

user
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Flame
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3 Answers3

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We can use the fundamental limit

$$\frac{\sin^2(x)}{x^2}=\left(\frac{\sin (x)}{x}\right)^2 \to 1$$

and since $\cos x$ is a continuous function

$$\lim_{x \to 0}\cos\left(\pi\cdot\frac{\sin^2(x)}{x^2}\right)=\cos (\pi \cdot 1)=-1$$

user
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  • hm I'll check the book once again if they are hinting towards one of those fundamental limits. For some reason I have never seen them before so I assumed that I had to rewrite this limit to a familiar form using trig identities. I see the other answers refer to the same fundamental limit as well. – Flame Sep 21 '20 at 22:09
  • @Flame Yes is is really foudamental. Refer to the link I've given for all the details to prove it. – user Sep 21 '20 at 22:15
  • Found it, there was a full page of it. They disguised it with a $\theta$ instead of an $x$ so my brain wired it to angles :) – Flame Sep 21 '20 at 22:38
  • @Flame Well done! This is an excellent reference for that! Bye – user Sep 21 '20 at 22:40
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    Minor clarification: Do you mean the fundamental limit? – chrylis -cautiouslyoptimistic- Sep 22 '20 at 07:37
  • Yes of course! I fix the typo. Thanks a lot – user Sep 22 '20 at 07:39
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What you need is the continuity of the cosine function. In particular, its continuity at $x=\pi$.

Let $\displaystyle g(x)=\pi\cdot \frac{\sin^2x}{x^2}$. Then $\lim_{x\to 0}g(x)=\pi$ using the fact that $$ \lim_{x\to 0}\frac{\sin x}{x}=1 $$ and the "multiplication law" of limits.

Now by continuity, $$ \lim_{x\to 0}\cos(g(x))=\cos(\lim_{x\to 0}g(x))=\cos(\pi)=-1. $$

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\begin{align*} \lim_{x \rightarrow 0} &{}\cos \left( \frac{\pi \sin^2 x}{x^2} \right) \\ &= \cos \left( \lim_{x \rightarrow 0}\frac{\pi \sin^2 x}{x^2} \right) & &\text{cosine is continuous} \\ &= \cos \left( \pi \lim_{x \rightarrow 0}\frac{\sin^2 x}{x^2} \right) & &\text{constant multiple} \\ &= \cos \left( \pi \left(\lim_{x \rightarrow 0} \frac{\sin x}{x}\right)^2 \right) & &\text{$x \mapsto x^2$ is continuous} \\ &= \cos \left( \pi \left(1\right)^2 \right) & &\text{trigonometric limit} \\ &= -1 \text{.} \end{align*}

The trigonometric limit used is one of a pair of fundamental limits, $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$ and $\lim_{x \rightarrow 0} \frac{1 - \cos x}{x} = 0$, which are usually found by application of the squeeze theorem.

Eric Towers
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