3
$\begingroup$

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?

$\endgroup$
2
6
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.