Finding the limit of $\frac {e^{-1/x^2}}{x^{100}}$ as $x \rightarrow 0$ 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? 
 A: 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 $50^{50}\cdot 0^{50}=0$.
A: 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 $\dfrac\infty\infty$
A: 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


*

*The Rapidity of the Exponential Function Towards Infinity
