lim_x->inf[sqrt(x^2+1)-x]

[ahorn]

Hi
I'd appreciate some help with this limit:
\(\displaystyle \lim_{x \to \infty}[\sqrt{x^2+1}-x]\)

So far, I've tried L'Hopital's Rule and re-ordering the equation:

\(\displaystyle \lim_{x \to \infty}\frac{x \sqrt{x^2+1}-1}{x}\)\(\displaystyle =\lim_{x \to \infty} \frac{\sqrt{x^2+1}+\frac{2x^2}{2 \sqrt{x^2+1}}}{1} \)\(\displaystyle =\lim_{x \to \infty}\frac{2x^2+1}{\sqrt{x^2+1}} \)\(\displaystyle = \lim_{x \to \infty} \frac{4x}{\frac{2x}{2\sqrt{x^2+1}}}\) etc.

 

[pka]

Hi
I'd appreciate some help with this limit:
\(\displaystyle \lim_{x \to \infty}[\sqrt{x^2+1}-x]\)

So far, I've tried L'Hopital's Rule

That does not work well.

\(\displaystyle \sqrt{x^2+1}-x=\dfrac{1}{\sqrt{x^2+1}+x}\)

 

[lookagain]

Hi
I'd appreciate some help with this limit:
\(\displaystyle \lim_{x \to \infty}[\sqrt{x^2+1}-x]\)

So far, I've tried L'Hopital's Rule and re-ordering the [expression]:

\(\displaystyle \displaystyle\lim_{x \to \infty}\frac{x \sqrt{x^2+1}-1}{x} \ \ \ \ \ \ \ \) <------ This is not correct. \(\displaystyle \ \)It should be \(\displaystyle \ \ \displaystyle\lim_{x \to \infty}\frac{x \sqrt{x^2 + 1} \ - \ x^2}{x} \)

.

 

[ahorn]

That does not work well.

\(\displaystyle \sqrt{x^2+1}-x=\dfrac{1}{\sqrt{x^2+1}+x}\)

Sorry, I should have tried multiplying by the conjugates. So the limit = 0?