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I need to calculate the limit or prove that the limit does not exists:

$$ \lim_{n \to \infty} \left ( 1 + \sin \frac{1}{n^2} \right )^{n^2} $$

Can someone guide how to solve this?

And please, don't just write the answer, it's my homework and I need to learn, not to copy.

Andronicus
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Alon
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    I know, but i didnt know that you can use that in proorfs, because its close to but not equal and we are talking about limits. have you seen proof in limits which uses this rule ? – Alon Jan 11 '20 at 13:07
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    It seems weird, like this:? $\displaystyle \lim_{n \to \infty} \left ( 1 + \sin \frac{1}{n^2} \right )^{n^2}$ = $\displaystyle \lim_{n \to \infty} \left ( 1 + \frac{1}{n^2} \right )^{n^2} = e$? that seems too simple, isn't it? – Alon Jan 11 '20 at 13:14
  • Ok thank you, i wish that you are right, if i use Andronicus's answer, its again, just copying. – Alon Jan 11 '20 at 13:17
  • Very similar question: What is the best method to solve the limit $\lim_{x\to \infty}\biggl(1+\sin\frac{2}{x^2}\biggr)^{x^2}$? Found using Approach0.See also: How to search on this site? – Martin Sleziak Jan 11 '20 at 13:17
  • The big difference is that i look for help to understand by myself, for a hint, and what you linked is a strait asnwer – Alon Jan 11 '20 at 13:18
  • @dxdydz: In general such replacements are invalid and can lead to incorrect answers. Don't assume things unless they are well described by some theorems. – Paramanand Singh Jan 12 '20 at 08:59
  • @dxdydz: you can easily construct examples yourself. But here is one generated on the fly. Since $(\sin x) /x\to 1$ as $x\to 0$ we can replace $\sin x $ by $x$ in scenarios where $x\to 0$ and therefore limit of $(x-\sin x) / x^3$ as $x\to 0$ is same as that of $(x-x) /x^3=0$ (this is incorrect). In general you can't replace $A$ by $B$ unless $A=B$. By the same logic the limit of $(x-\sin x) /\sin^3x$ is same as that of $(x-\sin x) / x^3$ as $x\to 0$ (this is correct). – Paramanand Singh Jan 12 '20 at 15:52
  • @dxdydz : In algebra we usually don't assume things which are not guaranteed by any theorems eg one never writes $\sqrt{a+b} =\sqrt{a} +\sqrt{b} $, but the trend in calculus is to assume whatever suits and luckily most of the time you get right answer. – Paramanand Singh Jan 12 '20 at 15:58

6 Answers6

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$$\lim_{n \to \infty}\left(1+\sin\left(\frac{1}{n}\right)\right)^{n}=\exp\left(\lim_{n \to \infty}\sin\left(\frac{1}{n}\right)n\right)=\exp\left(\lim_{n \to \infty}\frac{\sin\left(\frac{1}{n}\right)}{\frac{1}{n}}\right)=\exp\left(1\right)=e$$

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You can rewrite it as follows:

$$ \begin{align} &\lim_{n\rightarrow \infty}\left( 1+\sin \frac{1}{n^2} \right) ^{n^2}\\ =&\lim_{n\rightarrow \infty}\exp \left( n^2\ln \left( 1+\sin \frac{1}{n^2} \right) \right) \\ =&\lim_{n\rightarrow \infty}\exp \left( n^2\sin \frac{1}{n^2}\ln \left( \left( 1+\sin \frac{1}{n^2} \right) ^{\frac{1}{\sin \frac{1}{n^2}}} \right) \right) \\ =&\exp(1 \cdot1)=e\end{align}$$

Because $n^2 \sin\frac{1}{n^2}\to1$ and $\left( 1+\sin \frac{1}{n^2} \right) ^{\frac{1}{\sin \frac{1}{n^2}}}\rightarrow e$.

AsukaMinato
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Andronicus
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  • @Alon Notice, that I multiplied and divided by $\sin{\frac{1}{n^2}}$ and the other part was taken into logarithm, hence it's in the power – Andronicus Jan 11 '20 at 13:33
  • Yea i got that a min after i asked, thanks for the help – Alon Jan 11 '20 at 13:36
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The basic intuitive idea is that you should know that $(1+\frac{1}{n})^n \to e$, as $ n \to \infty$ and $\lim_{x \to 0}\frac{\sin(x)}{x}=1$.

So for large $n$ we can use $\frac{1}{n^2}$ instead of $\sin(\frac{1}{n^2})$ and so we are "really" looking at the limit of $(1+ \frac{1}{n^2})^{n^2}$ and using $m=n^2$, this is just really the limit of $(1+\frac1m)^m$ as $m$ grows big, so expect the limit to be $e$.

The rest is merely technical justification for this.

Henno Brandsma
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  • Good answer, really the only one that addresses the OP's plea in the last paragraph. – zhw. Jan 11 '20 at 17:11
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A classic trick to solve questions like this is to exploit the fact that $x \mapsto e^x$ is continuous, so: \begin{align*} \lim_{n \to \infty} \left(1 + \sin{\frac{1}{n^2}}\right)^{n^2} &= \lim_{n \to \infty} e^{n^2\ln\left(1 + \sin{\frac{1}{n^2}}\right)} \\ &= e^{\lim\limits_{n \to \infty} n^2\ln\left(1 + \sin{\frac{1}{n^2}}\right)} \end{align*} We can evaluate the limit in the exponent using L'Hopital Rule: \begin{align*} \lim_{n \to \infty} n^2\ln\left(1 + \sin{\frac{1}{n^2}}\right) &= \lim_{n \to \infty} \frac{\ln\left(1 + \sin{\frac{1}{n^2}}\right)}{\frac{1}{n^2}} \\ &= \lim_{n \to \infty} \frac{\frac{-\frac{2}{n^3}\cos{\frac{1}{n^2}}}{1 + \sin{\frac{1}{n^2}}}}{-\frac{2}{n^3}} \\ &= \lim_{n \to \infty} \frac{\cos{\frac{1}{n^2}}}{1 + \sin{\frac{1}{n^2}}} \\ &= 1 \end{align*} Therefore: $$ \lim_{n \to \infty} \left(1 + \sin{\frac{1}{n^2}}\right)^{n^2} = e^1 = e $$

Clement Yung
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A very simple way determines first the limit of the logarithm and use asymptotic equivalents:

Set $u=\frac1{n^2}$; you get $$ n^2\ln\Bigl(1+\sin\frac1{n^2}\Bigr)=\frac{\ln(1+\sin u)}{u}\sim_{u\to 0}\frac{\ln(1+u)}u, $$ and the latter tends to$1$ when $u\to 0$ by a well-known high-school limit, so by continuity of the exponential, $$\lim_{n\to\infty} \Bigl(1+\sin\frac 1{n^2}\Bigr)^{\!n^2}=\mathrm e^1=\mathrm e.$$

Bernard
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Yes this can be solved ! when n tends to infinity try inverting everything with using an h=1/x where x tends to 0 and you will be easily solving it for the power thing for such type of limits use logarithm .

Ravinder Dudeja
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