The sequence in question:
$$(u_{n})_{n\ge 0}=\sqrt{n}\left[\dfrac{e^{\sqrt{n+1}}}{e^{\sqrt{n-1}}}-1\right]$$
I see that there is an indeterminate form that must be lifted to calculate the limit, but I'm unsure how to start. How can this problem be approached?
Given limit:$$\lim_{x \to \infty} \sqrt{x}[e^{\sqrt{x+1}-\sqrt{x-1}}-1]$$ We can multiply both numerator and denominator by $\sqrt{x+1}-\sqrt{x-1}$ to get $$\lim_{x \to \infty} \sqrt{x}\frac{[e^{\sqrt{x+1}-\sqrt{x-1}}-1]}{\sqrt{x+1}-\sqrt{x-1}}[\sqrt{x+1}-\sqrt{x-1}]$$Since $\sqrt{x+1}-\sqrt{x-1} \to 0$ as $x \to \infty$ and we know that $$\lim_{x \to 0} \frac{e^x-1}{x}=1$$ So $$\lim_{x \to \infty} \frac{e^{\sqrt{x+1}-\sqrt{x-1}}-1}{\sqrt{x+1}-\sqrt{x-1}}=1$$ and the limit becomes$$\lim_{x \to \infty} \sqrt{x}[\sqrt{x+1}-\sqrt{x-1}]$$ After multiplying and dividing by $\sqrt{x+1}+\sqrt{x-1}$ and using identity $(a+b)(a-b)=a^2-b^2$, we obtain$$\lim_{x \to \infty}\frac{2\sqrt{x}}{\sqrt{x+1}+\sqrt{x-1}}$$ Dividing numerator and denonimator by $\sqrt{x}$ $$\lim_{x \to \infty}\frac{2}{\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}}$$ $$\lim_{x \to \infty}\frac{2}{\sqrt{1+0}+\sqrt{1-0}}=\frac{2}{2}=1$$