Limit problem. L'Hospital's Rule not allowed. [closed]
You can solve the limit without using a double-angle identity:
\begin{align*} \lim_{x\to\infty}\frac{\sin(2/x)+2/x}{\sin(1/x)}&=\lim_{t\to0}\frac{\sin(2t)+2t}{\sin(t)}\\ &=\lim_{t\to0}\left[\frac{\sin(2t)}{\sin(t)}+2\frac{t}{\sin(t)}\right]\\ &=\lim_{t\to0}\left[\frac{\sin(2t)}{1}\cdot\frac{1}{\sin(t)}+2\frac{t}{\sin(t)}\right]\\ &=\lim_{t\to0}\left[\frac{\sin(2t)}{2t}\cdot\frac{2t}{\sin(t)}+2\frac{t}{\sin(t)}\right]\\ &=\lim_{t\to0}\frac{\sin(2t)}{2t}\cdot2\lim_{t\to0}\frac{t}{\sin(t)}+2\lim_{t\to0}\frac{t}{\sin(t)}\\ &=1\cdot2+2\\ &=4 \end{align*}
There are two keys to this problem. I won't provide a full solution, but I will leave the tools below as hints:
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As @IvoTerek mentioned, $\lim\limits_{t\to 0}\frac{\sin(t)}{t} = 1$. So making the substitution $t=1/x$ gets you part of the way there.
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The other hint is that $\sin(2t)=2\sin(t)\cos(t)$. This identity will be needed to complete the problem.
Good luck!
Note that $$\lim_{x \to \infty} \frac{\sin({\frac{2}{x}})+\frac{2}{x}}{\sin({\frac{1}{x}})} = \lim_{y \to 0} \frac{\sin(2y)+2y}{\sin(y)} = \lim_{y \to 0} \frac{2\sin(y)\cos(y)+2y}{\sin(y)} \\= 2\lim_{y \to 0}\cos(y) + 2 \lim_{y \to 0}\frac{y}{\sin(y)}$$ We have $\lim_{y \to 0}\cos(y) = 1$ and $\lim_{y \to 0} \frac{\sin(y)}{y} = \lim_{y \to 0} \frac{\sin(y) - \sin(0)}{y-0} = \sin'(0) = \cos(0) = 1$, thus your limit is $4$.