# L'Hospital's rule doesn't converge for a function with square root

I was trying to find the limit of a function of the form $$\frac{x}{\sqrt{(x+a)(x+b)}}$$. When I apply L'Hospital's rule and differentiate the numerator and denominator, after simplification I end up with the same form as I started with:

$$L = \lim_{x \to \infty} \frac{x}{\sqrt{(x+a)(x+b)}} \\ = \lim_{x \to \infty} \frac{1}{\frac{1}{2} \frac{(2x+a+b)}{\sqrt{(x+a)(x+b)}}} = \frac{2\sqrt{(x+a)(x+b)}}{2x+a+b} = \frac{\frac{2x+a+b}{\sqrt{(x+a)(x+b)}}}{2} \\ = \lim_{x \to \infty} \frac{x}{\sqrt{(x+a)(x+b)}} \textrm{(because \lim_{x \to \infty} \frac{a+b}{\sqrt{...}} is zero)}$$

I noticed if I solve it by computing the limit of the squared value, the solution is easy: $$L^2 = \lim_{x \to \infty} \frac{x^2}{(x+a)(x+b)} \\ L^2 = \lim_{x \to \infty} \frac{2x}{2x+b+c} = 1 \\ L = \sqrt{1} = 1$$

Before I found this solution, I was searching online for different ways of evaluating limits and applying L'Hospital's rule, but none of the resources I came across seem to cover this case. Am I missing another straightforward way of solving this (whether using L'Hospital's or not)? When does L'Hospital's rule fail to converge, and what does it mean when it does?

• – Clement C. Jan 18 at 4:02
• L'Hospital's Rule does not fail here. If the new function converges to $L$ so does the old function. Only it does not help you to find the $L$ using differentiate and plug mantra. And that's because you can't plug $x=\infty$. – Paramanand Singh Jan 19 at 12:36

Discussion of L'Hopital's Rule aside, the straightforward way to solve the limit would be to multiply both numerator and denominator by $$\frac1x$$:
$$L = \lim_{x \to \infty} \frac{x}{\sqrt{(x+a)(x+b)}} = \lim_{x \to \infty} \frac{1}{\sqrt{\frac {(x+a)(x+b)}{x^2}}}=\lim_{x \to \infty} \frac{1}{\sqrt{\left(1+\frac ax\right)\left(1+\frac bx\right)}}=1$$