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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?

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    $\begingroup$ Relevant: en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule#Complications $\endgroup$
    – Clement C.
    Jan 18, 2021 at 4:02
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    $\begingroup$ 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 $x=\infty$. $\endgroup$
    – Paramanand Singh
    Jan 19, 2021 at 12:36
  • $\begingroup$ L'Hospital told you: If the limit of f(x) exists, then the limit of f(x) exists and is the same. Which is absolutely correct, just not useful. There are some clever functions around where this happens, but I haven't seen one yet where the limit couldn't be found easily in a different way. $\endgroup$
    – gnasher729
    Dec 28, 2023 at 12:55

1 Answer 1

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Discussion of L'Hopital's Rule aside, the straightforward way to solve the limit would be to multiply both numerator and denominator by $\frac1x$:

$$L = \lim_{x \to \infty} \frac{x}{\sqrt{(x+a)(x+b)}} = \lim_{x \to \infty} \frac{1}{\sqrt{\frac {(x+a)(x+b)}{x^2}}}=\lim_{x \to \infty} \frac{1}{\sqrt{\left(1+\frac ax\right)\left(1+\frac bx\right)}}=1$$

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