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Question Number 28467 by Asad8992002 last updated on 26/Jan/18

Commented by abdo imad last updated on 26/Jan/18

for all x from R and x>0        −(1/x)≤ ((sinx)/x)≤ (1/x) but  lim_(x→+∞)   −(1/x)=0  and  lim_(x→+∞)  (1/x)=0  so  lim_(x→+∞)  ((sinx)/x) =o    and by the same method we prove  lim_(x→−∞)    ((sinx)/x)=0  (  use the ch.x=−t).

$${for}\:{all}\:{x}\:{from}\:{R}\:{and}\:{x}>\mathrm{0}\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{{x}}\leqslant\:\frac{{sinx}}{{x}}\leqslant\:\frac{\mathrm{1}}{{x}}\:{but} \\ $$$${lim}_{{x}\rightarrow+\infty} \:\:−\frac{\mathrm{1}}{{x}}=\mathrm{0}\:\:{and}\:\:{lim}_{{x}\rightarrow+\infty} \:\frac{\mathrm{1}}{{x}}=\mathrm{0}\:\:{so} \\ $$$${lim}_{{x}\rightarrow+\infty} \:\frac{{sinx}}{{x}}\:={o}\:\:\:\:{and}\:{by}\:{the}\:{same}\:{method}\:{we}\:{prove} \\ $$$${lim}_{{x}\rightarrow−\infty} \:\:\:\frac{{sinx}}{{x}}=\mathrm{0}\:\:\left(\:\:{use}\:{the}\:{ch}.{x}=−{t}\right). \\ $$

Answered by malwaan last updated on 26/Jan/18

zero

$$\mathrm{zero} \\ $$

Answered by A1B1C1D1 last updated on 26/Jan/18

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