# 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?

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.
Do you know $$\lim_{x\to 0}\frac{\sin x}{x}=1?$$ Then try $$\lim_{x\to \infty}x\sin\frac{1}{x},$$ and eventually $$\lim_{x\to \infty}x^4\cdot x\sin\frac{1}{x}.$$
They use the result that $lim_{x \to 0}{\frac{sin(x)}{x}=1}$, or equivalently$lim_{x \to \infty}{\frac{sin(\frac{1}{x})}{\frac{1}{x}}=1}$.
You can rewrite your limit as $$lim_{x \to \infty}{x^4\frac{sin(\frac{1}{x})}{\frac{1}{x}}}$$. Therefore you can conlude
I see the solution containing $x^4 \frac{\sin(\frac{1}{x})}{\frac{1}{x}}$
If the solution uses this step, then you should probably know that: $$\lim_{x\to 0}\frac{\sin x}{x}=1$$ From this, it should be easy to see that also: $$\color{blue}{\lim_{x\to +\infty}\frac{\sin \tfrac{1}{x}}{\tfrac{1}{x}}=1}$$ So then: $$\lim_{x\to +\infty}\left(x^5\sin \tfrac{1}{x} \right)= \lim_{x\to +\infty} \left( x^4\frac{\sin \tfrac{1}{x}}{\tfrac{1}{x}} \right) = \lim_{x\to +\infty}x^4\color{blue}{\lim_{x\to +\infty}\frac{\sin \tfrac{1}{x}}{\tfrac{1}{x}}} = \cdots$$