how to find the limit of this question $\lim_{x\to +\infty}{\frac{\ln{(x+\sqrt{x^2+1})}-\ln{(x+\sqrt{x^2-1})}}{(e^{\frac{1}{x}}-1)^2}}$?

Question

$$\lim_{x\to +\infty}{\frac{\ln{(x+\sqrt{x^2+1})}-\ln{(x+\sqrt{x^2-1})}}{(e^{\frac{1}{x}}-1)^2}}$$

then I do a few transformation， and get $$\lim_{x\to +\infty}{x^2\frac{x+\sqrt{x^2+1}-x-\sqrt{x^2+1}}{x+\sqrt{x^2-1}}}$$ $$=\lim_{x\to +\infty}{x^2\frac{2}{x+\sqrt{x^2-1}}\frac{1}{\sqrt{x^2+1}+\sqrt{x^2-1}}}$$

but I don't know how to do the next step of this.

the correct answer is $\frac{1}{2}$

thanks for your help

Answer 1

ok, I just know how to solve this problem, thanks to Gary's tips

then I get $$x^2\frac{2}{4x^2}$$

and the answer is $\frac{1}{2}$

Answer 2

Spread the $x^2$ among the two numerators, then divide both fractions by $x:$

$$\lim_{x\to \infty}\frac{2}{1+\sqrt{1-x^{-2}}}\frac{1}{\sqrt{x^{-2}+1}+\sqrt{1-x^{-2}}} = 2/2 \cdot 1/2 = 1/2.$$

Check your steps, the numerator of the 2nd equation is $0$ as its currently written.

Answer 3

Set $\dfrac1x=h$

$$\ln(x+\sqrt{x^2+1})-\ln(x+\sqrt{x^2-1})=\ln\dfrac{x+\sqrt{x^2+1}}{x+\sqrt{x^2-1}}=\ln\left(1+\dfrac{\sqrt{1+h^2}-\sqrt{1-h^2}}{1+\sqrt{1-h^2}}\right)$$

We get

$$\dfrac{\lim_{h\to0^+}\dfrac{\ln\left(1+\dfrac{\sqrt{1+h^2}-\sqrt{1-h^2}}{1+\sqrt{1-h^2}}\right)}{\dfrac{\sqrt{1+h^2}-\sqrt{1-h^2}}{1+\sqrt{1-h^2}}}}{\left(\lim_{h\to0^+}\dfrac{e^h-1}h\right)^2}\cdot\lim_{h\to0^+}\dfrac{\dfrac{\sqrt{1+h^2}-\sqrt{1-h^2}}{1+\sqrt{1-h^2}}}{h^2}$$

Now use $\lim_{y\to0}\dfrac{\ln(1+y)}y=1$

and rationalize $\sqrt{1+h^2}-\sqrt{1-h^2}$