Find $\lim_{x\to\infty} \left(\cos\sqrt{\frac{2a}{x}}\right)^{x}$ where $a>0$

Question

Find $\lim_{x\to\infty} \left(\cos\sqrt{\frac{2a}{x}}\right)^{x}$ where $a>0$

I used L'Hopital and I reached to $$e^{\lim_{x\to\infty}\frac{-\sqrt{2a}}{2}x^{\frac{1}{2}}\tan\frac{\sqrt{2a}}{\sqrt{x}} }$$ what should I do now?

Answer 1

When $x\to \infty$, $\tan\frac{\sqrt{2a}}{\sqrt{x}}\sim \frac{\sqrt{2a}}{\sqrt{x}}$. So $$e^{\lim_{x\to\infty}\frac{-\sqrt{2a}}{2}x^{\frac{1}{2}}\tan\frac{\sqrt{2a}}{\sqrt{x}} }=e^{\lim_{x\to\infty}\frac{-\sqrt{2a}}{2}x^{\frac{1}{2}}\frac{\sqrt{2a}}{\sqrt{x}} }=e^{-a}.$$

Answer 2

Alternatively: $$\begin{align}\lim_{x\to\infty} \left(\cos\sqrt{\frac{2a}{x}}\right)^{x}&=\lim_{x\to\infty} \left(\left[1+\cos\sqrt{\frac{2a}{x}}-1\right]^{\frac{1}{\cos\sqrt{\frac{2a}{x}}-1}}\right)^{\frac{\cos\sqrt{\frac{2a}{x}}-1}{\frac 1x}}=\\ \exp\left(\lim_\limits{x\to\infty}\frac{\cos\sqrt{\frac{2a}{x}}-1}{\frac 1x} \right)&\overbrace{=}^{LH}\exp\left(\lim_\limits{x\to\infty}\frac{-\sin{\sqrt{\frac{2a}{x}}}\cdot \sqrt{2a}\cdot \frac{-x^{-3/2}}{2}}{-x^{-2}}\right)=\\ \exp\left(\lim_\limits{x\to\infty} \frac{\sin{\sqrt{\frac{2a}{x}}}}{\sqrt{\frac{2a}{x}}}\cdot \lim_\limits{x\to\infty} \frac{-2a}{2}\right)&=\exp(-a).\end{align}$$

Answer 3

Better have used Taylor series. If so we have$$\cos\sqrt{\dfrac{2a}{x}}=1-\dfrac{a}{x}+o(\dfrac{a}{x})$$which means that the limit is $$e^{-a}$$