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

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