Evaluating limits approaching infinity

Question

How to calculate the limit of this function?

$$f(x) = \frac{(-1)^x \sqrt{x-1}}{x},\ x = \{1, 2, 3, 4...\}$$

So, I have tried calculating the limits way:

First, by using the multiplication rule;

$$\begin{equation} \begin{split} \lim_{x\to+\infty}\frac{(-1)^x}{x}.\lim_{x\to+\infty}\sqrt{x-1} \end{split} \end{equation}$$

And then, I used the sandwich theorem to evaluate the limit of the function on the left.

$$ \lim_{x\to+\infty} \frac{(-1)^x}{x}=0\quad \text{because}\ -\frac{1}{x} \leq \frac{(-1)^x}{x} \leq \frac{1}{x} $$

Both the upper and lower boundary functions have a limit of $\ 0$, as $x\to+\infty$.

Evaluating the function on the right:

$$\lim_{x\to+\infty} \sqrt{x-1} = \infty$$

I don't know if this is right or wrong?

$$\lim_{x\to+\infty} f(x) = 0\cdot\infty = 0$$

Answer 1

We can say that for $x\in\mathbb{N}$ $$\left\lvert\frac{(-1)^x \sqrt{x-1}}{x}\right\rvert \le \frac{\sqrt{x}}{x}=\frac{1}{\sqrt x}.$$ Now take the limit as $x\to+\infty$ and the expression equals zero. Finally $\lvert \lim_{x\to+\infty}f(x) \rvert\le0\Rightarrow \lim_{x\to+\infty}f(x)=0$.

Answer 2

You cannot do this. $\lim fg = \lim f \times \lim g $ only if both the limits of $f$ and $g$ exist, but in your case $\lim_{x\to \infty} \sqrt{x-1} $ doesn’t exist.

Instead you can say something like $$\left | \frac{(-1)^x \sqrt{x-1}}{x} \right |=\frac{\sqrt{x-1}}{x}=\sqrt{ \frac{1}{x} -\frac{1}{x^2}}\to \sqrt{0-0} =0 $$