Limit of $\lim_{x \to +\infty} x \left( \sqrt{(1+\frac{a}{x}) (1+\frac{h}{x})} -1 \right)$

Question

Is there a way to calculate this limit?

$$\lim_{x \to +\infty} x \left( \sqrt{(1+\frac{a}{x}) (1+\frac{h}{x})} -1 \right)$$

I know it's equal to $\frac{a+h}{2}$, but what method can I use to calculate this? I can't use l'Hopital.

Answer 1

Hint:

$$x\left(\sqrt{\left(1+\frac ax\right)\left(1+\frac hx\right)}-1\right)=\left(\sqrt{\left(x+a\right)\left(x+h\right)}-x\right)=\frac{{\left(x+a\right)\left(x+h\right)}-x^2}{\sqrt{\left(x+a\right)\left(x+h\right)}+x}.$$

Answer 2

We have that

$$\lim_{x \to +\infty} x \left( \sqrt{(1+\frac{a}{x}) (1+\frac{h}{x})} -1 \right)=\lim_{x \to +\infty}\frac{ \sqrt{(1+\frac{a}{x}) (1+\frac{h}{x})} -1 }{\frac1x}$$

then use derivative definition by $y=\frac1x\to 0$ and

$$f(y)=\sqrt{(1+ay)(1+by)} \implies f'(y)=\frac12\frac{a(1+by)+b(1+ay)}{\sqrt{(1+ay)(1+by)}}$$

Answer 3

$$\lim_{x \to +\infty} x \left( \sqrt{(1+\frac{a}{x}) (1+\frac{h}{x})} -1 \right) =$$ $$= \lim_{x \to +\infty} x \left( \sqrt{(1+\frac{a}{x}) (1+\frac{h}{x})} -1 \right) \cdot \frac{\sqrt{(1+a/x)(1+h/x)}+1}{\sqrt{(1+a/x)(1+h/x)}+1} =$$ $$=\lim_{x\to+\infty}\frac{x}{\sqrt{(1+a/x)(1+h/x)}+1}\left((1+a/x)(1+h/x)-1\right) = \lim_{x\to+\infty} \frac{x((a+h)/x+ah/x^2)}{\sqrt{(1+a/x)(1+h/x)}+1} =$$ $$= \lim_{x\to+\infty} \frac{(a+h)+ah/x}{\sqrt{(1+a/x)(1+h/x)}+1} = \frac{a+h}{2}$$