Question: Determine $ \ \large \lim\_{x \to \infty} \frac{40-2^{-0.1x+5} \sin (x)}{5-\frac{1}{x}} \ $

**Answer:**

$ \ \large \lim\_{x \to \infty} \frac{40-2^{-0.1x+5} \sin (x)}{5} \\ = \ \large \lim\_{x \to \infty} \frac{40}{5-\frac{1}{x}}- \ \large \lim\_{x \to \infty} \frac{2^{-0.1x+5} \sin (x)}{5-\frac{1}{x}} \\ =8 - \frac{0 \cdot \lim\_{x \to \infty} \sin (x}{5-0} \\ =8-0 \\ =8$

**Am I right?**

Answer: 
See my comments above. You are almost correct. Better argue as follows. Note that
as $x\to +\infty$ we have $$0\leq |2^{-0.1x+5}\sin x|\leq 2^{-0.1x+5}\to 0.$$
This implies (using the Squeeze Theorem) that $$\lim\_{x\to +\infty} 2^{-0.1x+5}\sin x=0$$
Yes, you are right:
$$ \lim\_{x \to +\infty} \frac{40-2^{-0.1x+5} \sin (x)}{5-\frac{1}{x}}=\frac{40-\lim\limits\_{x\rightarrow+\infty}\frac{32\sin{x}}{2^{0.1x}}}{5-\lim\limits\_{x\rightarrow+\infty}\frac{1}{x}}=8$$