Limit where l'Hopital's rule doesn't help

[kostas]

Homework Statement

<br /> \lim_{x\to +\infty} \frac{x}{\sqrt{x^2+1}} <br />

Homework Equations

The Attempt at a Solution

<br /> \lim_{x\to +\infty} \frac{x}{\sqrt{x^2+1}} \stackrel{l&#039;H}{=} \lim_{x\to +\infty} \frac{\sqrt{x^2+1}}{x} \stackrel{l&#039;H}{=} \lim_{x\to +\infty} \frac{x}{\sqrt{x^2+1}}<br />

I get \frac{+\infty}{+\infty} and l'H rule doesn't seem to help. Is the above correct? If yes, what else could I try?

 

[Defennder]

One way is to use a trigo substitution. Another more intuitive way is to note that x^2 + 1 tends to simply x^2 for large x.

 

[chislam]

There is another way that you can find the limit as x goes to infinity. If you just divide each term by the highest power of x, so in this case it would simply be x (and not x^2 because it's within a square root so we treat it as being x). Then you can split the limit up between terms and you can find the limit easily.

Here's an example.

<br /> \lim_{x\to\infty} \frac{3x^3}{5x^3 + x^2 + 6}<br />

Now if you divide each term by x^3 you get:

<br /> \lim_{x\to\infty} \frac{3}{5 + 1/x + 6/x^3}<br />

So eventually the 1/x and 6/x^3 will "disappear" and the limit would be 3/5.

 

[NoMoreExams]

To add to what the other 2 posters have said, note that you can factor x^2 inside the square root and the fact that \sqrt{x^{2}} = |x| should lead you to the answer.

 

[jgens]

Adding more to what's already been said: The application of L'Hospital's rule repeatedly yields the previous function's mulitplicative inverse. Therefore, if the limit exists, the limit and it's multiplicative inverse must be the same. What does that suggest about the limit?

 

[kostas]

Thank you all!