limit of $4x\sin(\frac1x)$

Question

I'm having a bit of a problem taking the limit of functions involving $\sin(\dfrac1x)$.

Mainly, I don't know whether or not L'Hopital's rule is required. Here's the particular problem:

$$\lim\limits_{x\to \infty} 4x\sin (\frac1x)= ?$$

Answer 1

$$ \lim_{x\to \infty}\left[4x\sin\left(\frac1x\right)\right]= 4\lim_{x\to \infty}\frac{ \sin\left(\frac1x\right)}{\frac1x} $$

As $x\to\infty$, $\frac1x \to0$. Substitute $\frac1x$ with a new variable ($\theta=\frac1x$):

$$ 4\lim_{\theta\to 0}\frac{ \sin\theta}{\theta}=4\cdot 1=4. $$

$\displaystyle\lim_{x\to0}\frac{\sin x}{x}=1$ is a well-known limit and the fact that it's equal to 1 is proven in elementary calculus.

Or you can use L'Hospital's rule (the indeterminate form is $\frac00$):

$$4\lim_{x\to\infty}\frac{\sin\left(\frac1x\right)}{\frac1x}\stackrel{\text{L'H}}{=} 4\lim_{x\to\infty}\frac{\left[\sin\left(\frac1x\right)\right]'}{\left(\frac1x\right)'}=4\lim_{x\to\infty}\frac{\cos{\left(\frac1x\right)}\left(-\frac{1}{x^2}\right)}{-\frac{1}{x^2}}=\\ 4\lim_{x\to\infty}\cos{\left(\frac1x\right)}=4\cdot\cos{0}=4\cdot 1=4. $$

Answer 2

Let $h=1/x$. As $x\to \infty$, $h\to 0$.

$$\lim_{x\to \infty}4x\sin\left(\dfrac{1}{x}\right)=4\cdot\lim_{x\to \infty}\dfrac{\sin\left(1/x\right)}{1/x}\mapsto4\cdot\underbrace{\lim_{h\to 0}\dfrac{\sin h}{h}}_{\to 1}\to 4$$