Answer to Question #218292 in Calculus for Nkululeko

Question #218292
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)
1
Expert's answer
2021-07-19T05:50:08-0400

Suppose x0.x\not=0.


0sin(ex)x1x,xR,x00\leq|\dfrac{\sin(e^x)}{x}|\leq \dfrac{1}{|x|}, x\in\R, x\not=0

limx1x=0\lim\limits_{x\to\infin}\dfrac{1}{|x|}=0

Then by the Squeeze Theorem


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

0sin(ex)x2+2sin(ex)x,xR,x00\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


limxsin(ex)x2+2=0\lim\limits_{x\to\infin}|\dfrac{\sin(e^x)}{\sqrt{x^2+2}}|=0

Therefore


limxsin(ex)x2+2=0\lim\limits_{x\to\infin}\dfrac{\sin(e^x)}{\sqrt{x^2+2}}=0


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