How to Evaluate $\lim_{x \to \infty} 2 + 2x\sin\left(\frac{4}{x}\right)$? Here is my limit to be evaluated

$\lim_{x \to \infty} 2 + 2x\sin\left(\frac{4}{x}\right)$=?

 A: Here is a simple step by step approch
$$\begin{align}
&\quad \lim_{x \to \infty} 2 + 2x\sin\left(\frac{4}{x}\right) \\
&= \lim_{x \to \infty} 2 + 8 \frac{x}{4} \sin\left(\frac{4}{x}\right) \\
&= \lim_{x \to \infty} 2 + 8 \lim_{x \to \infty} \frac{\sin\left(\frac{4}{x}\right)}{\frac{4}{x}} \\
&= \lim_{x \to \infty} 2 + 8 \lim_{u \to 0} \frac{\sin(u)}{u} \\
&=2+8(1)=10
\end{align} 
$$
A: BIG HINT:
$$\lim_{x\to\infty}\left(2+2x\sin\left(\frac{4}{x}\right)\right)=2+\left(\lim_{x\to\infty}2x\sin\left(\frac{4}{x}\right)\right)=2+2\left(\lim_{x\to\infty}x\sin\left(\frac{4}{x}\right)\right)=$$

Let $n=\frac{1}{x}$:

$$2+2\left(\lim_{n\to 0}\frac{1}{n}\cdot\sin\left(4n\right)\right)=2+2\left(\lim_{n\to 0}\frac{\sin\left(4n\right)}{n}\right)=$$
$$2+2\left(\lim_{n\to 0}\frac{\frac{\text{d}}{\text{d}n}\left(\sin\left(4n\right)\right)}{\frac{\text{d}}{\text{d}n}\left(n\right)}\right)=2+2\left(\lim_{n\to 0}\frac{4\cos(4n)}{1}\right)=2+8\lim_{n\to 0}\cos(4n)$$
A: The limit in question is $2+2\lim_{x\to\infty}x\sin\frac{4}{x}=2+8=10$.
A: For all $n$$$2+2x\sin (\frac nx)=2+2 \frac{\sin (\frac nx)}{\frac 1x}=2+2n \frac{\sin (\frac nx)}{\frac nx}\rightarrow 2+2n$$
