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Finding the limit of: $$\frac {e^{-1/x^2}}{x^{100}}$$ as $x \rightarrow 0$

My answer is:

1- I can not apply L`hopital rule because the limit will still be 0/0 because the numerator will still be exponential and the power of the denominator is increasing .... am I correct?

2- I will solve it by taking into account the exponential growth relative to the polynomial growth and knowing that the exponential growth is much faster than the polynomial growth then the exponential term is the dominant term and the limit will be zero .... am I correct ? .... and will my answer deserve a full credit?

Intuition
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3 Answers3

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You should notice that this boils down to the limits $$\lim_{x\to \infty}\frac{\log x} {x} =0\text{ or }\lim_{x\to\infty} \frac{x} {e^x} =0$$ You can use L'Hospital's Rule to prove the above limits. There are proofs based on simpler tools like Squeeze Theorem.

For the current question just put $1/x^2=t$ to get $$\lim_{t\to \infty }\frac{t^{50}}{e^t}$$ Further if we put $t=50u$ then the above limit gets transformed into $$\lim_{u\to\infty}50^{50} \left(\frac{u}{e^u}\right)^{50}$$ and this equals $50^{50}\cdot 0^{50}=0$.

Paramanand Singh
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Set $1/x^2=y$ to find $$\lim_{x\to0}\dfrac{e^{-1/x^2}}{x^{100}}=\lim_{y\to\infty}\dfrac{y^{50}}{e^y}$$

Now apply L'Hospital as the ratio is of the from $\dfrac\infty\infty$

lab bhattacharjee
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  • I think it will take a huge number of times of applying l`hopital also .... am I correct?.....what about my second question ? – Intuition Nov 13 '18 at 09:03
  • @hopefully, Yes you can take $50$th derivative or expand $e^y$ – lab bhattacharjee Nov 13 '18 at 09:05
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    @hopefully You are right with your your second question. In this cases exponential functions always "win" against polinomial functions and those always "win" again logarythmic functions. If you will get full credit depends on your corrector and how detailed you explain your argument (and why you are allowed to use it). – Kaligule Nov 13 '18 at 09:24
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As noticed we can also apply l’Hopital several times or as an alternative we have that

$$\frac {e^{-1/x^2}}{x^{100}}=\frac {1}{x^{100}e^{1/x^2}}\to 0$$

indeed for all $y>0$ and $n$ eventually

$$e^y\ge y^n \implies e^{1/x^2}\ge \frac1{x^{102}}$$

and thus

$$\frac{1}{x^{100}e^{1/x^2}}\le \frac{1}{x^{100}\frac1{x^{102}}}=x^2\to 0$$

Refer also to the related

user
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  • Could you describe more your forth line ....... your $y = -1/x^2$ ? and n .... what? – Intuition Nov 13 '18 at 09:54
  • For any $y>0$ and exponent $n>0$ we have that eventually $e^y\ge y^n$, now take $y=1/x^2$ and $n=51$ to obtain the RHS. – user Nov 13 '18 at 09:57
  • @hopefully Since the exponential growth is greater than any polynomial growth (the fact is obviuosly true as an extension also for $n$ real). – user Nov 13 '18 at 09:58
  • @hopefully If we assume the result as a given the limit follows immediately otherwise we need to prove that. – user Nov 13 '18 at 09:59
  • are you assuming that $x$ is positive? – Intuition Nov 13 '18 at 10:12
  • @hopefully Since in that case $\frac {e^{-1/x^2}}{x^{100}}=\frac {e^{-1/x^2}}{{(x^2)}^{50}}$ we can wlog assume $x>0$. Note indeed that I've ssumed $y=1/x^2 >0$ which is always true for $x\neq 0$. – user Nov 13 '18 at 10:16
  • Once we provethat eventually $e^y\ge y^n$ for any $y>0$ and $n>0$ the result follows. – user Nov 13 '18 at 10:18
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    @hopefully I've added a reference for that proof. – user Nov 13 '18 at 10:27