I have the function $$f(x)=\frac{1}{2^x}-3x+2, \quad x\in\mathbb{R}$$ and I want to find the $$\lim_{x\to -\infty}{\frac{\sin(f(x))}{x}}.$$ I have that $$\lim_{x\to -\infty}{f(x)}=+\infty,$$ so I am thinking of using the squeeze theorem to find $$\lim_{x\to -\infty}{\frac{\sin(f(x))}{x}},$$ but this wasn't work for me. Any ideas? (I don't want to use de L'Hospital Rule)

  • $\begingroup$ Hint: $\sin$ is bounded. $\endgroup$ – Michael Burr Feb 6 '18 at 14:10
  • $\begingroup$ Start with $-1 \leq \sin(f(x)) \leq 1$. $\endgroup$ – Jonathan Feb 6 '18 at 14:11

Yes simply squeeze theorem noting that

$$\left|\frac{\sin(f(x))}{x}\right| \le \left|\frac{1}{x}\right|$$

and this directly leads to $0$.


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