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Verify $$\lim_{x\to\infty} \frac {7 \sin x}{\sqrt{5x}}$$

I want to say that the answer is D.N.E. because of sin.

If anyone can tell me if I am right and help explain why it would be D.N.E. I would be grateful.

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I don't think it is DNE because it oscilates but each oscilation is smaller than the previous one. I would say the limit is $0$. –  chubakueno Sep 8 '13 at 22:01

2 Answers 2

up vote 11 down vote accepted

Hint: $$ -\frac {7}{\sqrt{5x}}\le \frac {7\sin{x}}{\sqrt{5x}} \le \frac {7}{\sqrt{5x}} $$

If you do not know the squeeze theorem,

Note that the LHS and RHS tend to $0$ as $x\to \infty$, squeeze theorem tells us the middle must also tend to $0$, as $x\to \infty$

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...and sandwich theorem for completion. –  AlexR Sep 8 '13 at 22:05
    
@Rustyn-Yazdanpour ok so the LHS is $-\infty$ and the RHS is $\infty$ so I have to find the middle hence my answer will be 0 –  John Beal Sep 8 '13 at 22:22
    
Both side tends to zero, because x is in the denominator. –  Stefan4024 Sep 8 '13 at 22:23

The limit is clearly $0$. Your numerator oscillates perpetually between $\pm 7$, while your denominator diverges to $+\infty$. While squeeze theorem is instructive, demonstrative, and du rigeur, you might want to ponder the reason I used the word "clearly" in evaluating a limit. Your considering the possibility of D.N.E. is my only concern here.

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