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I find difficulty proving the no existence of this limit

I show my process

$$ \lim_{x\to 0} \biggl(1 + x e^{- \frac{1}{x^2}}+\sin \frac{1}{x^4}\biggr)^{e^{\frac{1}{x^2}}}$$

We begin with rewriting the limit as follows: $$ \lim_{x\to 0} \biggl(1 + x e^{- \frac{1}{x^2}}+\sin \frac{1}{x^4}\biggr)^{e^{\frac{1}{x^2}}}=\lim_{x\to 0} \biggl( 1 + x \frac{1}{e^{\frac{1}{x^2}}} +\sin \frac{1}{x^4}\biggr)^{e^{\frac{1}{x^2}}}$$ and analyze the various addends and the exponent of the limit: $$\begin{align*} &x \frac{1}{e^{\frac{1}{x^2}}}\to 0\\ &\sin \frac{1}{x^4}\to \not \exists\\ &e^{\frac{1}{x^2}}\to+\infty.\\ \end{align*}$$ The problem here lies in the fact that we have an addendum that there is no limit, let's consider: $$\sin{a_n}\quad\text{e}\quad\sin{b_n}\quad\text{con }\quad n\to+\infty$$ where the two sequences are: $$a_n=\frac{\pi}{2}+2n\pi\quad\text{e}\quad b_n=2n\pi$$ Then, the function values ​​calculated in the sequence $ a_n $, with $ k $ positive integer, tends to $ 1 $, calculated values ​​of the sequence $b_n$ tends to $0$,: $$\lim_ {n \to\infty}\sin{a_n}=1 \quad\text{and}\quad \lim_{n\to \infty}\sin {b_n} = 0$$ and therefore, as we know, the limit of $\sin x$ ($x \to \infty$) not exists.

Now, to prove that the given limit does not exist,i continued in this way

$t= \frac{1}{x^2},$ (if $x\to0 \rightarrow t\to+\infty$) : $$\lim_{x\to 0} \biggl( 1 + x \frac{1}{e^{\frac{1}{x^2}}} +\sin \frac{1}{x^4}\biggr)^{e^{\frac{1}{x^2}}}=\lim_{t\to +\infty} \biggl( 1 + \frac{ \sqrt{t}}{t}\cdot \frac{1}{e^t} +\sin{t^2}\biggr)^{e^t}$$

$\frac{ \sqrt{t}}{t}\cdot \frac{1}{e^t}\to 0$

i consider $$\begin{align*} &\lim_{t\to +\infty} \biggl(1 + \sin{({a_n})^2}\biggr)^{e^t}=\biggl(1+1\biggr)^{e^t}=+\infty\\ &\lim_{t\to +\infty} \biggl( 1 + \sin{({b_n})^2}\biggr)^{e^t}=e^{e^t\ln\biggl( 1 + \sin{({b_n})^2}\biggr)}=e^{+\infty\ln( 1 + 0)}=??? \end{align*}$$

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up vote 1 down vote accepted

Given an integer $k$, let $\frac{1}{x^4}=2\pi k + \frac{\pi}{2}$, or $x = (2\pi k + \frac{\pi}{2})^{-\frac{1}{4}}$. Then $\sin {\frac{1}{x^4}} = 1$. Let $y=e^{\frac{1}{x^2}}$. Then your expression is:


Now, if $|x|<1$ then $y>e$, so $\frac{x}{y}> \frac{-1}{2}$. So this expression is at least as big as $(\frac{3}2)^y$.

Picking a large enough $k$, we can make $x$ arbitrarily small, so $\frac{1}{x^2}$ arbitrarily large, so $y=e^{\frac{1}{x^2}}$ arbitrarily large. So, we see that we can make your expression bigger than $(\frac{3}2)^Y$ for arbitrarily large $Y$, and hence it has no limit.

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That's half of it, but I think the OP also wants to prove that the limit isn't $+\infty$. – Lopsy Dec 29 '11 at 21:31

Hint: Better consider the subsequence where $\sin(x)=-1$ instead of the one where $\sin(x)=0$, it is easy to see that the limit is $0$ there and then you found a subsequence that converges to $0$ and one that converges to $+\infty$ (the one where $\sin(x)=1$). Hence the limit is not existent.

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for example $b_n=\frac{3\pi}{2}+2n\pi$? – FrConnection Dec 29 '11 at 21:29
Exactly, but note that you have to consider $b_n^{1/4}$ as values for $t$ :) – Listing Dec 29 '11 at 21:32
do you mean $b_n= (2\pi n)^{1/4}$? – FrConnection Dec 29 '11 at 21:41
I mean you have to take the root because you don't want $\sin(b_n)=-1$ but $\sin(b_n^4)=-1$, you get $b_n^{1/4}=(\frac{3\pi}{2}+2n\pi)^{1/4}$ – Listing Dec 29 '11 at 21:56
yes! sorry! clear! – FrConnection Dec 29 '11 at 22:07

You can show that $$\lim_{t\to +\infty} \biggl( 1 + \frac{ \sqrt{t}}{t}\cdot \frac{1}{e^t} +\sin{t^2}\biggr)^{e^t}$$ does not exist by observing that for arbitrarily large $t_0$, there is a $t>t_0$ such that $\sin t^2 = 1$ so the value at $t'$ is at least $2^{e^{t}}$, hence we have divergence.

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I think he also wants to exclude divergence towards infinity. – Listing Dec 29 '11 at 21:29
Ah. Well in that case this gives a divergent subsequence, and we also have points where $\sin t^2 = -1$, so we get a subsequence which converges to $0$ as well. – Alex Becker Dec 29 '11 at 21:33

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