How do I make a change of variable for $\;\lim_{x\to 0}\frac{\sqrt{x^2+p^2}-p}{\sqrt{x^2+q^2}-q}$?

Question

I can't use l'hopital, so change of variable is the only way. $$\lim_{x\to 0}\frac{\sqrt{x^2+p^2}-p}{\sqrt{x^2+q^2}-q}$$

Answer 1

We have: $\dfrac{\sqrt{x^2+p^2}-p}{\sqrt{x^2+q^2}-q}=\dfrac{\dfrac{x^2}{\sqrt{x^2+p^2}+p}}{\dfrac{x^2}{\sqrt{x^2+q^2}+q}}=\dfrac{\sqrt{x^2+q^2}+q}{\sqrt{x^2+p^2+p}}\to \dfrac{2q}{2p} = \dfrac{q}{p}$ as $x \to 0$.

Answer 2

We can do

$$\frac{\sqrt{x^2+p^2}-p}{\sqrt{x^2+q^2}-q} \times \frac{\sqrt{x^2+q^2}+q}{\sqrt{x^2+q^2}+q} = \frac{(\sqrt{x^2+p^2}-p)(\sqrt{x^2 + q^2}+q}{x^2}$$

First $$\frac{\frac{\sqrt{x^2+p^2}-p}{1}\times \sqrt{x^2 + q^2} +q}{x^2} \tag{1}$$

and

$$\frac{\frac{\sqrt{x^2 + p^2}+p}{\sqrt{x^2 + p^2}+p}\times \frac{\sqrt{x^2+p^2}-p}{1}\times \sqrt{x^2 + q^2} +q}{x^2} \tag{2}$$

and finally

$$\frac{\frac{x^2}{\sqrt{x^2 + p^2}+p}\times \sqrt{x^2 + q^2} +q}{x^2} = \frac{\frac{x^2( \sqrt{x^2 + q^2}+q)}{\sqrt{x^2 + p^2}+p}}{x^2} \tag{3}$$

to get

$$\frac{x^2(\sqrt{x^2 + q^2}+q)}{x^2 (\sqrt{x^2 + p^2}+p)} = \frac{\sqrt{x^2 + q^2}+q}{\sqrt{x^2 + p^2}+p}$$

Answer 3

You can write: $$\sqrt{x^2+p^2}-p=p\cdot\left(\sqrt{1+\frac{x^2}{p^2}}-1\right)$$ And: $$\sqrt{x^2+q^2}-q=q\cdot\left(\sqrt{1+\frac{x^2}{q^2}}-1\right)$$

We know that $(1+t)^\alpha-1 \,\, \sim\,\, \alpha \cdot t$ when $x\to 0$; notice that in our cases $t=\frac{x^2}{p^2}$ for the numerator and $t =\frac{x^2}{q^2}$ for the denominator.

So, we have: $$\lim_{x\to 0}\frac{\sqrt{x^2+p^2}-p}{\sqrt{x^2+q^2}-q}=\lim_{x\to 0}\frac{p\cdot\left(\sqrt{1+\frac{x^2}{p^2}}-1\right)}{q\cdot\left(\sqrt{1+\frac{x^2}{q^2}}-1\right)}\,\,\sim\,\,\lim_{x\to 0}\frac{p\cdot \frac{1}{2}\frac{x^2}{p^2}}{q\cdot \frac{1}{2}\frac{x^2}{q^2}}=\frac{q}{p}$$