# Limit of type 0/0 without L'Hopital

I am trying to figure out the limit as $h \rightarrow 0$ of

$$\frac{x\sqrt{x+h+1}-x\sqrt{x+1}}{h(x+h)}$$

without using l'Hopital's rule.

I have tried to extend the fraction by the conjugate of the numerator. It gets pretty hairy and didn't get me anywhere (so I will spare everyone having to read through my attempt). But basically the problem is I can't seem to get rid of $h$ in the denominator.

• Multiply the top and bottom by $x\sqrt{x+h+1}+x\sqrt{x+1}$ – JasonM Jun 14 '16 at 6:33
• Remember that the "conjugate factor" times the factor already in the numerator is intended to give you a "difference of two squares". So you do not need to multiply everything out. The numerator will be $$\ x \ (\sqrt{x+h+1} \ - \sqrt{x+1}) \ (\sqrt{x+h+1} \ + \sqrt{x+1}) \ \ = \ \ \ x \ [ \ (x+h+1) \ - \ (x+1 ) \ ] \ \ ,$$ which will simplify to leave you with the $\ h \$ in the numerator to cancel the one in the denominator. (Don't forget about the conjugate factor now also in the denominator, to which you can just apply the limit.) – colormegone Jun 14 '16 at 6:34
• "Pretty hairy": I wonder how. Multiplying by the conjugate results in five terms at the numerator, which reduce to a single one. Then after simplification of $h$, indeterminacy is gone. – Yves Daoust Jun 14 '16 at 6:39
$$\lim_{h \to 0}\frac{x\sqrt{x+h+1}-x\sqrt{x+1}}{h(x+h)}=\lim_{h \to 0}\frac{x}{x+h}\frac{\sqrt{x+h+1}-\sqrt{x+1}}{h}\tag{1}$$ let $f(x)=\sqrt{x+1}$ and $f'(x)=\frac{f(x+h)-f(x)}{h}$ then $(1)$ can be written as: $$f'(x)$$, so now differentiate $f(x)$.
A possible solution uses Taylor expansion built around $h=0$ $$\sqrt{x+h+1}=\sqrt{x+1}+\frac{h}{2 \sqrt{x+1}}-\frac{h^2}{8 (x+1)^{3/2}}+O\left(h^3\right)$$ So $$x\sqrt{x+h+1}-x\sqrt{x+1}=\frac{h x}{2 \sqrt{x+1}}-\frac{h^2 x}{8 (x+1)^{3/2}}+O\left(h^3\right)$$ The denominator being $hx+h^2$, long division leads to $$\frac{x\sqrt{x+h+1}-x\sqrt{x+1}}{h(x+h)}=\frac{1}{2 \sqrt{x+1}}-\frac{(5 x+4)}{8 x (x+1)^{3/2}}h+O\left(h^2\right)$$ which shows the limit and also how it is approached.