2

I would like to calculate

$$ \lim_ {n \to \infty} {\cos \left(\frac {a}{n}\right)^{n^2}} $$ where $n \in \mathbb N$ and $a \in \mathbb R \setminus \{0\}$

The answer should be $e^{-a^2/2}$, but I'm not sure how to calculate it.

user35603
  • 3,002
aprilduck
  • 147
  • Hint: Let $\cos(a/n)=1-x_n$. What do you know about the behaviour of $x_n$ when $n\to\infty$? – Did Dec 31 '15 at 12:38

3 Answers3

2

Notice, let $\frac 1n=t\implies t\to 0$ as $n\to \infty$ $$\lim_{n\to \infty}\left(\cos\left(\frac{a}{n}\right)\right)^{n^2}=\lim_{t\to 0}\left(\cos\left(at\right)\right)^{1/t^2}$$

$$=\exp\lim_{t\to 0}\frac{1}{t^2}\ln\left(\cos\left(at\right)\right)$$ $$=\exp\lim_{t\to 0}\frac{\underbrace{\ln\left(\cos\left(at\right)\right)}_{\longrightarrow 0}}{\underbrace{t^2}_{\longrightarrow 0}}$$ Using L'Hospital's rule for $\frac 00$ form $$=\exp\lim_{t\to 0}\frac{\frac{d}{dt}\left(\ln\left(\cos\left(at\right)\right)\right)}{\frac{d}{dt}(t^2)}$$

$$=\exp\lim_{t\to 0}\frac{\frac{-a\sin (at)}{\cos(at)}}{2t}$$ $$=\exp\left(\frac{(-a^2)}{2}\lim_{t\to 0}\left(\frac{\sin(at)}{at}\right)\cdot\lim_{t\to 0} \frac{1}{\cos (at)}\right)$$ $$=\exp\left(\frac{(-a^2)}{2}\left(1\right)\cdot(1)\right)$$ $$=\color{red}{e^{-a^2/2}}$$

Harish Chandra Rajpoot
  • 37,450
1

HINT:

$$\left(\cos\dfrac an\right)^{n^2}=\left(1-\sin^2\dfrac an\right)^{\dfrac{n^2}2}$$

$$=\left(\left(1-\sin^2\dfrac an\right)^{-\dfrac1{\sin^2\dfrac an}}\right)^{-\sin^2\dfrac an\cdot \dfrac{n^2}2}$$

As $n\to\infty,\dfrac an\to0\implies\sin^2\dfrac an\to0$ so the inner limit converges to $e$

Set $\dfrac an=h,$

$$\lim_{n\to\infty}\sin^2\dfrac an\cdot \dfrac{n^2}2=\dfrac{a^2}2\lim_{h\to0}\dfrac{\sin^2h}{h^2}=?$$

lab bhattacharjee
  • 274,582
  • 1
    Isn't there a gap in the argument, after "the inner limit converges to $e$"? This does not imply immediately that the overall limit is $e^{\lim_{n\to\infty} [...]}$... for instance, $1+\frac{1}{n}$ converges to $1$, but $(1+\frac{1}{n})^n$ goes to $e$ (not $1^{\lim n}=1$). – Clement C. Dec 31 '15 at 13:20
  • @ClementC., It work as the limit in the exponent is also non-zero & finite – lab bhattacharjee Dec 31 '15 at 13:24
  • 1
    I know — I am merely saying that explicitly stating so is important, as otherwise mimicking the argument can lead to errors. – Clement C. Dec 31 '15 at 13:26
1

Using Taylor series:

  • First, rewrite with exponentials: $$ \left(\cos\frac{a}{n}\right)^{n^2} = e^{n^2\ln \cos\frac{a}{n}} $$

  • Then, expand, composing $\cos u = 1-\frac{u^2}{2} + o(u^2)$ when $u\to0$ and $\ln(1+u)=u+o(u)$: (place your mouse over the shaded area to show the derivation)

    $$\left(\cos\frac{a}{n}\right)^{n^2} = e^{n^2\ln \left(1-\frac{a^2}{2n^2} +o(\frac{1}{n^2})\right)}= e^{n^2\left(-\frac{a^2}{2n^2} +o(\frac{1}{n^2})\right)}=e^{-\frac{a^2}{2} +o(1)} \xrightarrow[n\to\infty]{} e^{-\frac{a^2}{2}}$$

where the very last step relies on the fact that by continuity of the exponential, $e^{o(1)}\xrightarrow[n\to\infty]{} e^{0}=1$.

Clement C.
  • 67,323