real analysis - What is $lim_{x rightarrow infty}{x^5 sin(frac{1}{x})}$

What is $\lim_{x \rightarrow \infty}{x^5 \sin(\frac{1}{x})}$

I'm reading a solution to a problem and I don't understand. I see the solution containing $x^4 \frac{\sin(\frac{1}{x})}{\frac{1}{x}}$. Could someone please explain how to compute this limit?

Answer

The solution that you're reading probably works as follows. First of all, we can rewrite the given limit as
$$\lim\limits_{x\to\infty}x^5\sin\left(\frac{1}{x}\right)=\lim\limits_{x\to\infty}x^4\cdot x\sin\left(\frac{1}{x}\right)=\lim\limits_{x\to\infty}x^4\cdot\lim\limits_{x\to\infty}x\sin\left(\frac{1}{x}\right).$$
If we now focus on the second limit:

$$\lim\limits_{x\to\infty}x\sin\left(\frac{1}{x}\right)=\lim\limits_{x\to\infty}\frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}}=1,$$
according to the special limit $\lim\limits_{t\to0}\frac{\sin t}{t}=1$. Textbooks always use $x$ here, but I'm using $t$ instead to avoid confusion with $x$ of the original question. Just substitute $t=1/x$ to see that you have an instance of this limit rule.