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I need to find limit of $\lim_{n\to\infty}(\cos\frac{1}{n})^{n^{2}}$.

I tried going like this:

Let $x = \frac{1}{n}$,

$\lim_{n\to\infty}(\cos\frac{1}{n})^{n^{2}} = \lim_{x\to0^{+}}(\cos(x))^{1/x^2} = \lim_{x\to0^{+}}e^{\ln(\cos(x))^{1/x^2}}= \lim_{x\to0^{+}}e^{\frac{1}{x^2}\ln(\cos(x))}$

The answer is $1$, therefore I need to show that $\lim_{x\to0^{+}}e^{\frac{1}{x^2}\ln(\cos(x))}$ = $e^{0} = 1$

Meaning, I need to show that $\lim_{x\to0^{+}}\frac{1}{x^2}\ln(\cos(x)) = 0$.

I know that $\lim_{x\to0^{+}}\cos(x) = 1$, therefore, $\ln(1) =0$,

so I'm stuck at $\lim_{x\to0^{+}}e^{\frac{1}{0^+}0} = ? $

If I could say that $\frac{1}{0^+}0$ is $0$ that would solve it but I guess I can't.

Whats a better way to solve it?

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    $\cos x = 1 - \frac{1}{2} x^2 + O(x^4)$. Hence $\ln (\cos x) = -\frac{1}{2}x^2 + O(x^4)$. Who says the answer is $1$? – 2017-01-13
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    The calculator I use – 2017-01-13
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    Does it take the arguments of trigonometric functions in degrees? If we take $\bigl(\cos \frac{1}{n}°\bigr)^{n^2}$, the limit would be $\exp \bigl(-\frac{\pi^2}{64800}\bigr)$, and that might be just close enough to $1$ that the limited precision of the calculator can't distinguish it. – 2017-01-13
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    It is set to Radians. When giving $n=10^7$ or above as input, the result is 1 – 2017-01-13
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    Okay. That's another effect of limited precision, though. $\cos \bigl(10^{-7}\bigr) \approx 1 - 5\cdot 10^{-15}$, and many calculators have not enough precision to distinguish that number from $1$. If you compute $1 - \cos \bigl(10^{-7}\bigr)$, does your calculator give $0$ as the answer, or some very small positive value? If the latter, which? – 2017-01-13
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    It gives $0$ for that – 2017-01-13

3 Answers 3

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What is needed is that, as $x \to 0$, $\cos(x) \approx 1-x^2/2 $.

Then $(\cos(1/n))^{n^2} \approx (1-1/(2n^2))^{n^2} \to e^{-1/2} $ since $(1-1/n)^n \to 1/e$ as $n \to \infty$.

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L'Hopital's rule can help.

$$\lim_{x \to 0} \frac{\ln(\cos x)}{x^2} = -\lim_{x \to 0} \frac{\tan x}{2x} = -\lim_{x\to 0} \frac{1}{2}\sec^2 x = -\frac{1}{2}.$$

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First of all, your limit is wrong; your calculator probably can't handle such large numbers. It should be $\frac{1}{\sqrt{e}}$, as you can see empirically.

So with that in mind, by the same logic you used, we're looking to show

$$\lim_{x\to0^+}\frac{\ln(\cos x)}{x^2} = -\frac{1}{2}$$

This is easy to do with l'Hôpital's rule, but you don't want to use that. You could also use Taylor series, but usually the people who want an answer without l'Hôpital's rule don't want to use that, either. So we'll do it this way.

$$\lim_{x\to0^+}\frac{\ln(\cos x)}{x^2} =\lim_{x\to0^+}\left(\frac{\ln(1 + (\cos x - 1))}{\cos x-1}\cdot\frac{\cos x - 1}{x^2}\right)$$

This can be shown to be equal to $-\frac{1}{2}$ using the familiar limits $\lim_{x\to0}\frac{\ln(1+x)}{x} = 1$ and $\lim_{x\to0}\frac{1 - \cos x}{x^2} = \frac{1}{2}$.

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    For the "familiar limits," I know a [geometric proof](http://math.stackexchange.com/a/36341/43949) of the second limit. How do you prove the first limit? – 2017-01-14
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    $$\lim_{x\to0}\frac{\ln(1+x)}{x} = \lim_{x\to0}\ln((1+x)^{\frac{1}{x}}) = \ln e = 1$$ – 2017-01-14
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    Is it trivial that $\lim_{x\to0^+}\frac{\ln(1 + (\cos x - 1))}{\cos x-1} = \lim_{x\to0^+}(\frac{\ln(1+x)}{x})$? – 2017-01-14
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    It'd actually be $\lim_{x\to0^-}\frac{\ln(1+x)}{x}$, but beyond that, yeah, it's just a substitution, $u = \cos x - 1$. – 2017-01-14