# What if L'hospital's rule doesn't work?

Oct 5, 2014

l'Hopital's Rule occationally fails by falling into a never ending cycle. Let us look at the following limit.

${\lim}_{n \to \infty} \frac{\sqrt{{x}^{2} + 1}}{x}$

by l'Hopital's Rule ($\infty$/$\infty$),

$= {\lim}_{n \to \infty} \frac{\frac{x}{\sqrt{{x}^{2} + 1}}}{1} = {\lim}_{n \to \infty} \frac{x}{\sqrt{{x}^{2} + 1}}$

by l'Hopital's Rule ($\infty$/$\infty$),

$= {\lim}_{n \to \infty} \frac{1}{\frac{x}{\sqrt{{x}^{2} + 1}}} = {\lim}_{n \to \infty} \frac{\sqrt{{x}^{2} + 1}}{x}$

As you can see, the limit came back to the original limit after applying l'Hopital's Rule twice, which means that it will never yield a conclusion. So, we just need to try another approach.

${\lim}_{n \to \infty} \frac{\sqrt{{x}^{2} + 1}}{x}$

by including the denominator under the square-root,

$= {\lim}_{n \to \infty} \sqrt{\frac{{x}^{2} + 1}{x} ^ 2}$

by simplifying the expression inside the square-root,

${\lim}_{n \to \infty} \sqrt{1 + \frac{1}{x} ^ 2} = \sqrt{1 + \frac{1}{\infty} ^ 2} = \sqrt{1 + 0} = 1$

So, we could come up with the limit without using l'Hopital's Rule.

I hope that this was helpful.