2
$\begingroup$

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

3 Answers 3

0

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.

  • 0
    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.2011-12-29
2

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.

  • 0
    yes! sorry! clear!2011-12-29
1

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:

$(2+\frac{x}{y})^y$.

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.

  • 1
    That's half of it, but I think the OP also wants to prove that the limit isn't $+\infty$.2011-12-29