# How do I solve the following limit using L'Hospital's Rule?

I want to show that:

$\lim_{y\ -> 0+}$ e^(-1/y)/y^2 = 0.

Substituting directly we get an indeterminate limit of the form 0/0, so applying L'Hopital's Rule I get that the above limit is equal to:

$\lim_{y\ -> 0+}$ (e^(-1/y)/y^2)/2y,

which is also an indeterminate of the form 0/0.

Applying L'Hopital's Rule again I get:

$\lim_{y\ -> 0+}$ (e^(-1/y)/y^4-2e^(-1/y)/y^3)/2

I can't seem to get any further.

• You mean limit as $y \to 0$ of $\frac{e^-\frac{1}{y}}{y^2}$? Commented Aug 15, 2015 at 18:34
Let $\displaystyle L = \lim_{x\rightarrow 0^{+}}\frac{e^{-\frac{1}{x}}}{x^2}\;,$ Now put $\displaystyle x=\frac{1}{y}\;,$ Then $y\rightarrow \infty$ and $\displaystyle L = \lim_{y\rightarrow \infty}\frac{y^2}{e^y}$
Now Using $\bf{L,Hopital\; Rule}\;,$ we get $\displaystyle \lim_{y\rightarrow\infty}\frac{2y}{e^y}$
again Using $\bf{L,Hopital\; Rule}\;,$ we get $\displaystyle \lim_{y\rightarrow\infty}\frac{2}{e^y} = 0$
Bcz Here $\lim_{y\rightarrow \infty}e^y\rightarrow \infty$