Solving limit without applying L'Hospital's rule There's a practice problem I'm utterly stumped at. No matter what I try, I can't arrive at the correct answer. Please, if someone could help me understand, I would be very grateful:
$$ \lim_{x\to \infty}(x+3)\cdot \sin(\frac{1}{x})$$
 A: Assume we know $1 = \lim_{x\to 0} \sin(x)/x$.
We know that the limit of $\sin \left(\frac 1 x\right)$ exists, and thus
$$
\lim_{x\to\infty}(x+3)\cdot\sin\left(\frac 1 x\right) = \lim_{x\to\infty} x\cdot \sin\left(\frac 1 x\right) + 3\lim_{x\to\infty} \sin \left(\frac 1 x\right) = 1 + 0 = 1
$$
A: Hint:
Use equivalents: near $\infty$, one has
$$\sin\frac1x\sim \frac1x,\quad\text{so }\quad (x+3)\sin\frac1x\sim \frac{x+3}x.$$
A: You need two well-known limits to solve this one:
$$\lim_{x\to 0}\frac{\sin{x}}{x}=1$$
and
$$\lim_{x\to \infty}\sin{\left(\frac{1}{x}\right)}=0$$
When $x$ approaches infinity, $\frac{1}{x}$ approaches $0$. Therefore, $\lim_{x\to 0}\frac{\sin{x}}{x}$ is equivalent to $\lim_{x\to \infty}\frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}}$:
$$
\begin{align}
\lim_{x\to \infty}\left[(x+3)\cdot \sin\left(\frac{1}{x}\right)\right]
&=\lim_{x\to \infty}\left[x\cdot\sin\left(\frac{1}{x}\right)\right]+\lim_{x\to \infty}\left[3\cdot\sin\left(\frac{1}{x}\right)\right]\\
&=\lim_{x\to \infty}\frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}}+3\cdot\lim_{x\to \infty}\sin\left(\frac{1}{x}\right)\\
&=1+3\cdot0\\
&=1
\end{align}
$$
