# L'Hopital's Rule with $\lim \limits_{x \to \infty}\frac{2^x}{e^\left(x^2\right)}$

(a) Show that $$\lim \limits_{x \to \infty}\frac{2^x}{e^\left(x^2\right)}$$ is a standard indeterminate form, but that L'Hopital's Rule does not give you any information about the limit.

(b) Show that by taking the natural logarithm, however, we arrive at a limit that we can now evaluate. What is the value of the original limit?

So my approach to part (a) was that $$\lim \limits_{x \to \infty}\frac{2^x}{e^\left(x^2\right)} = \frac{\infty}{\infty}$$

which is an indeterminate form. By using L'Hopital's Rule and differentiating both numerator and denominator i get $$\lim \limits_{x \to \infty}\frac{2^x}{e^\left(x^2\right)} = \lim \limits_{x \to \infty}\frac{2^x \ln 2}{2e^\left(x^2\right)x} = \frac{\infty}{\infty}$$

So clearly L'Hopitals Rule doesnt work.

My approach to part (b):

Taking the natural logarithm I get $$\lim \limits_{x \to \infty}\frac{x \ln 2}{x^2} = \lim \limits_{x \to \infty}\frac{\ln 2}{x}= 0$$

Is my evaluation of part b correct? and is my answer to a reasonable?

Sorry if this is kind of a yes or no question, but it's the only way I could phrase it.

• One thing that's worth emphasizing in the first part is that not only does L'Hopitals rule not work after one use, it doesn't work after any number of iterations since you'll always end up something like $2^x/(\text{polynomial}\times e^{x^2})$. Commented Jul 29, 2014 at 19:49
• Also it is easy to notice that for $x$ large enough, we have $\exp(x^2)>2^{n x}$ for any $n>0$ (with the size of $x$ depending on $n$) Commented Feb 11, 2020 at 16:24
• Shame it doesn't say $+\infty$, then $$\lim_{x\to+\infty}\frac{2^x}{e^{x^2}}=\lim_{x\to+\infty}\frac{2^x}{\left(e^x\right)^x}=\lim_{x\to+\infty}\left(\frac2{e^x}\right)^x=0$$ because $2<e^x$ as $x\to+\infty$. Commented Aug 7, 2020 at 18:51

$\log(\dfrac{2^x}{e^{x^2}}) = \log(2^x) - \log(e^{x^2}) = x\log(2) - x^2 \to -\infty$ when $x\to \infty$, which means $\dfrac{2^x}{e^{x^2}} \to 0$