Is the limit of $\cos \frac {1}{2(\sqrt{x +1} - \sqrt {x})} $as $x \rightarrow \infty$ is 0?

Question

Is the limit of $\cos \frac {1}{2(\sqrt{x +1} - \sqrt {x})} $as $x \rightarrow \infty$ is 0?

My confusion is:

The denominator is undetermined quantity $\infty - \infty.$

If so how can I proof this by epsilon delta definition if my function is $$\cos \frac {1}{2(\sqrt{x +1} - \sqrt {x})} \sin \frac {1}{2(\sqrt{x +1} + \sqrt {x})} $$

I want to prove that the limit is 0, asI am sure thatthe limit of the sine function is 0.

Answer 1

The second function (with the product) surely goes to $0$ because $|\cos (.) | \leq 1$. The first function (the one in the title) does not tend to $0$. To see this note that for eacn $n$ there exists $x_n$ such that $\frac {\sqrt {x_{n+1}}+\sqrt {x_n}} 2 =\frac {(2n+1) \pi }2$. This can be seen from the fact that $\frac {\sqrt {x+1}+\sqrt {x}} 2$ increases continuously from $1$ to $\infty$ on $(0,\infty)$. It follows that the given function $f$ satisfies $|f(x_n)|=1$ with $x_n \to \infty$.

Answer 2

$$\frac{1}{\sqrt{x+1}-\sqrt{x}}=\sqrt{x+1}+\sqrt{x},$$ which says that the limit does not exist.

Indeed, for all natural $n$ there is unique $x_n$, for which $$\sqrt{x_n+1}+\sqrt{x_n}=\pi+2\pi n.$$ Thus, $$\lim_{n\rightarrow+\infty}\cos\frac{1}{2(\sqrt{x_n+1}-\sqrt{x_n})}=\lim_{n\rightarrow+\infty}\cos\left(\frac{\pi}{2}+\pi n\right)=0.$$ Also, it's obvious that $\lim\limits_{n\rightarrow+\infty} x_n=+\infty.$

By the same way we can show that there is $y_n$ such that $y_n\rightarrow+\infty$ for which $$\lim_{n\rightarrow+\infty}\cos\frac{1}{2(\sqrt{y_n+1}-\sqrt{y_n})}=1.$$ Id est, $$\lim_{x\rightarrow+\infty}\cos\frac{1}{2(\sqrt{x+1}-\sqrt{x})}$$ does not exist.