Determine $ \ \large \lim_{x \to \infty} \frac{40-2^{-0.1x+5} \sin (x)}{5-\frac{1}{x}} \ $ 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?

Yes, you are right:
$$ \lim_{x \to +\infty} \frac{40-2^{-0.1x+5} \sin (x)}{5-\frac{1}{x}}=\frac{40-\lim\limits_{x\rightarrow+\infty}\frac{32\sin{x}}{2^{0.1x}}}{5-\lim\limits_{x\rightarrow+\infty}\frac{1}{x}}=8$$
