Answer to Question #218292 in Calculus for Nkululeko

Question

use squeeze theorem to show that: limit as x approaches infinity of (sin(e^x))/x =0, hense evaluate the limit as x approaches infinity of (sin(e^x))/sqr(x^2+2)

Answer 1

Suppose $x\not=0.$

$$0\leq|\dfrac{\sin(e^x)}{x}|\leq \dfrac{1}{|x|}, x\in\R, x\not=0$$

$$\lim\limits_{x\to\infin}\dfrac{1}{|x|}=0$$

Then by the Squeeze Theorem

$$\lim\limits_{x\to\infin}|\dfrac{\sin(e^x)}{x}|=0$$

$$0\leq|\dfrac{\sin(e^x)}{\sqrt{x^2+2}}|\leq |\dfrac{\sin(e^x)}{x}|, x\in\R, x\not=0$$

Then by the Squeeze Theorem

$$\lim\limits_{x\to\infin}|\dfrac{\sin(e^x)}{\sqrt{x^2+2}}|=0$$

Therefore

$$\lim\limits_{x\to\infin}\dfrac{\sin(e^x)}{\sqrt{x^2+2}}=0$$