Find the limit of $\lim_{x\to 0}\frac{\sqrt{x^2+a^2}-a}{\sqrt{x^2+b^2}-b}$

Question

Can someone help me solve this limit?

$$\lim_{x\to0}\frac{\sqrt{x^2+a^2}-a}{\sqrt{x^2+b^2}-b}$$

with $a>0$ and $b>0$.

Answer 1

Without using L'Hopital $$\lim_{x\to0}\frac{\sqrt{x^2+a^2}-a}{\sqrt{x^2+b^2}-b}\cdot\frac{\sqrt{x^2+a^2}+a}{\sqrt{x^2+b^2}+b}\cdot\frac{\sqrt{x^2+b^2}+b}{\sqrt{x^2+a^2}+a}$$

Answer 2

No need for L'Hopital - we simply multiply and divide by the conjugate radical expression: \begin{align} \frac{\sqrt{x^2+a^2}-a}{\sqrt{x^2+b^2}-b}&=\left(\frac{\sqrt{x^2+a^2}-a}{\sqrt{x^2+b^2}-b}\cdot\frac{\sqrt{x^2+a^2}+a}{\sqrt{x^2+b^2}+b}\right)\cdot\frac{\sqrt{x^2+b^2}+b}{\sqrt{x^2+a^2}+a} \\ &=\frac{x^2+a^2-a^2}{x^2+b^2-b^2}\cdot\frac{\sqrt{x^2+b^2}+b}{\sqrt{x^2+a^2}+a}= \frac{\sqrt{x^2+b^2}+b}{\sqrt{x^2+a^2}+a}\to\frac{2b}{2a}=\frac{b}{a}. \end{align}

Answer 3

Without using L'Hopital, expand the root in a power series, to get: $$\sqrt{x^2+a^2} -a = \frac{x^2}{2a}+O(x^4)$$ Leading to the limit of the ratio being $b/a$.