A limit involving irrational powers: can I use L'Hopital?

Question

I was trying to evaluate the following limit: $$\lim_{x\to\infty}\frac{(x^{\sqrt 2}+1)^{\sqrt 2}}{x^2+1}$$ My first instinct was to use L'Hopital, but it just makes things messier (because of the irrational power), and after applying it four times, I gave up on it.

How else might I evaluate this limit?

Answer 1

\begin{align} \lim_{x\to\infty}\frac{(x^{\sqrt 2}+1)^{\sqrt 2}}{x^2+1} &=\lim_{x\to\infty}\frac{x^2\left(1+x^{-\sqrt 2}\right)^{\sqrt 2}}{x^2 \left(1+\frac{1}{x^2}\right)}\\ & =\lim_{x\to\infty}\frac{\left(1+\frac 1{x^{\sqrt 2}}\right)^{\sqrt 2}}{\left(1+\frac{1}{x^2}\right)}\\ &=1 \end{align}

Can you see how?

Answer 2

You shouldn't need to apply L'hopital's 4 times; after two applications the denominator will be a constant function and it should be easy to evaluate.

Answer 3

Set $x=1/t$ and compute instead $$ \lim_{t\to0^+} \frac{\left(\dfrac{1}{\mathstrut t^{\sqrt{2}}}+1\right)^{\!\sqrt{2}}}{\dfrac{1}{t^2}+1} =\lim_{t\to0^+}\frac{(1+t^{\sqrt{2}})^{\sqrt{2}}}{1+t^2} $$