Answer to Question #201494 in Calculus for Sanjeshni

Question #201494

Use the Squeezing Theorem to show that

lim

X approaches to +sum to infinity

Sin x divide by x =0


1
Expert's answer
2021-06-03T03:20:40-0400
1sinx1-1\leq\sin x\leq 1

If x>0x>0


1xsinxx1x-\dfrac{1}{x}\leq\dfrac{\sin x}{x}\leq \dfrac{1}{x}

limx+(1x)=0 and limx+(1x)=0\lim\limits_{x\to+ \infin}(-\dfrac{1}{x})=0\ and\ \lim\limits_{x\to +\infin}(\dfrac{1}{x})=0

Therefore by the Squeezing Theorem


limx+(sinxx)=0\lim\limits_{x\to + \infin}(\dfrac{\sin x}{x})=0



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