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Does anyone know how to evaluate the following limit?

$$\lim_{x\to\infty}\left(\cos\frac{3}{x}\right)^{x^2}$$

I want to see a step by step solution if possible, Thank you.

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closed as off-topic by Jyrki Lahtonen, Mark Fantini, apnorton, Yiorgos S. Smyrlis, Ivo Terek Sep 20 '14 at 15:56

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    $\begingroup$ Taylor expansion. Taking the logarithm is optional. $\endgroup$ – Daniel Fischer Sep 20 '14 at 13:14
  • $\begingroup$ $\cos\dfrac3x = 1 - \dfrac{(3/x)^2}2 + \text{higher-degree terms}$, and $\left(1 + \dfrac{-9/2}{x^2}\right)^{x^2}\to e^{-9/2}$ as $x^2\to\infty$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Sep 20 '14 at 13:16
  • $\begingroup$ @MichaelHardy: You should post that as an answer. $\endgroup$ – Robin Goodfellow Sep 20 '14 at 13:19
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    $\begingroup$ A general trick to handle limits of the type $1^\infty$ is the following result: If $a(x)\to1$, $b(x)\to\infty$ and $b(x)(a(x)-1)\to A$, then $$a(x)^{b(x)}\to e^A.$$ $\endgroup$ – Jyrki Lahtonen Sep 20 '14 at 13:42
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rewrite the limit in the form $e^{\lim_{x \to\infty}\frac{\ln\left(\cos\left(\frac{3}{x}\right)\right)}{\frac{1}{x^2}}}$ and use the rules of L'Hospital. The searched result is $e^{-9/2}$
Sonnhard.

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    $\begingroup$ Maybe use $\exp(x)$ instead of $e^x$ since that fraction in a limit in a power isn't very readable. $\endgroup$ – Alice Ryhl Sep 22 '14 at 14:44
  • $\begingroup$ but it is the most usefull idea to compute such limits $\endgroup$ – Dr. Sonnhard Graubner Sep 22 '14 at 14:46
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    $\begingroup$ I agree with your solution, I simply propose using a different notation for $e^x$ namely $\exp(x)$ $\endgroup$ – Alice Ryhl Sep 22 '14 at 14:47
  • $\begingroup$ Legibility: $\exp\left(\lim\limits_{x\to\infty}\dfrac{\ln(\cos(3/x))}{1/x^2}\right)$ ${}\qquad{}$ $\endgroup$ – Michael Hardy Sep 22 '14 at 15:14
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Recall that $$ \left( 1 + \frac a r \right)^r \to e^a \text{ as }r\to\infty. $$ and $$ \cos x = 1 - \frac{x^2}2 + \frac{x^4}{24} - \frac{x^6}{720} + \cdots. $$ and in particular $$ \cos\frac 3 x = 1 - \frac{(3/x)^2}{2} + \text{higher-degree terms}. $$

So we have $$ \left(1 + \dfrac{-9/2}{x^2}\right)^{x^2}\to e^{-9/2}. $$ I posted this as a comment rather than as an answer because I didn't want to write out how to deal with the higher-degree terms. But, goaded on by Robin Goodfellow's comment, I'm putting it here.

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$\cos\dfrac3x=1-2\sin^2\dfrac3{2x},$

$$\lim_{x\to\infty}\left(\cos\frac3x\right)^{x^2}=\left(\lim_{x\to\infty}\left[1+\left(-2\sin^2\dfrac3{2x}\right)\right]^{--2\sin^2\dfrac3{2x}}\right)^{\left(\lim_{x\to\infty}-2x^2\sin^2\dfrac3{2x}\right)}$$

The inner limit $=e$

For the exponent set $\dfrac3{2x}=h\iff x=\dfrac3{2h}$

$$\lim_{x\to\infty}-2x^2\sin^2\dfrac3{2x}=-2\cdot\dfrac94\left(\lim_{h\to0}\dfrac{\sin h}h\right)^2$$

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$$ \lim\limits_{x\to \infty} \left(\cos\frac{3}{x}\right)^{x^2}$$ $$= \lim\limits_{x\to \infty} \exp\left(\ln\left(\cos\frac{3}{x}\right)^{x^2}\right)$$ $$= \lim\limits_{x\to \infty} \exp\left(x^2\ln\left(\cos\frac{3}{x}\right)\right)$$ Let $t=\frac{3}{x}$, then $$\lim\limits_{t\to 0} \exp\left(\frac{9\ln\left(\cos t\right)}{t^2}\right)$$ $$=\exp\left(9\lim\limits_{t\to 0}\frac{\ln\left(\cos t\right)}{t^2}\right)$$ $$=\exp\left(9\lim\limits_{t\to 0}\frac{\cos t-1}{t^2}\right)$$ $$=\exp\left(-9\lim\limits_{t\to 0}\frac{1-\cos t}{t^2}\right)$$ $$=\exp\left(-\frac92\lim\limits_{t\to 0}\frac{t^2}{t^2}\right)$$ $$=\exp\left(-\frac92\right)$$

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