# Infinite limit, small question

1. May 21, 2010

### Asphyxiated

1. The problem statement, all variables and given/known data

$$\lim_{x \to \infty} \frac {4x}{\sqrt{2x^{2}+1}}$$

2. Relevant equations

3. The attempt at a solution

So at first glance I saw that this equation ends up at infinity/infinity so I tried to use L'hopital's rule and got:

$$\lim_{x \to \infty} \frac {4}{2x(2x^{2}+1)^{-1/2}}$$

That seemed to have sent me in a wrong direction so I went back and just factored out the x^2 from the radical and it worked out such as:

$$\lim_{x \to \infty} \frac {4x}{\sqrt{x^{2}(2+1x^{-2})}}$$

$$\lim_{x \to \infty} \frac {4x}{x\sqrt{2+1x^{-2}}}$$

$$\lim_{x \to \infty} \frac {4}{\sqrt{2+1x^{-2}}}$$

and with the limit applied:

$$\lim_{x \to \infty} \frac {4}{\sqrt{2+1x^{-2}}} = \frac {4}{\sqrt{2}} = \sqrt{8}$$

which is correct, so my question is why didn't L'Hopital's Rule work when it was infinity/infinity or would it have worked out and I just stopped too soon?

2. May 21, 2010

### Staff: Mentor

It's MUCH simpler to factor x^2 out of both terms in the radical, which is where you eventually went.
The above can be written as
$$\lim_{x \to \infty} \frac {4\sqrt{2x^2 + 1}}{2x}}$$

This is still the indeterminate form [inf/inf], and I think that repeated applications of L'Hopital's Rule won't be of any use.