MHB Limit of a trigonometric function

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The discussion focuses on evaluating three specific limits involving trigonometric functions. The limits in question are \(\lim_{x\rightarrow \infty }x\cdot sin(x)\), \(\lim_{x\rightarrow 0 }\frac{sin(x)}{\sqrt{x}}\), and \(\lim_{x\rightarrow \infty }\frac{sin(x)}{x}\). A hint suggests that the third limit approaches 0 due to the behavior of a bounded function multiplied by a function approaching 0. However, it is clarified that the first limit does not approach 0 because neither function tends to 0, and the oscillatory nature of the sine function indicates that it does not have an infinite limit. The discussion emphasizes the relationships between these limits and the known limit \(\lim_{x\rightarrow 0 }\frac{sin(x)}{x}\).
Yankel
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Hello all,

I need some guidance in solving these limits:

\[\lim_{x\rightarrow \infty }x\cdot sin(x)\]

\[\lim_{x\rightarrow 0 }\frac{sin(x)}{\sqrt{x}}\]

\[\lim_{x\rightarrow \infty }\frac{sin(x)}{x}\]

I guess that the second and third ones are somehow related to

\[\lim_{x\rightarrow 0 }\frac{sin(x)}{x}\]

But I am not sure how to convert the known limit into the new ones.

Thank you in advance.
 
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Yankel said:
Hello all,

I need some guidance in solving these limits:

\[\lim_{x\rightarrow \infty }x\cdot sin(x)\]

\[\lim_{x\rightarrow 0 }\frac{sin(x)}{\sqrt{x}}\]

\[\lim_{x\rightarrow \infty }\frac{sin(x)}{x}\]

I guess that the second and third ones are somehow related to

\[\lim_{x\rightarrow 0 }\frac{sin(x)}{x}\]

But I am not sure how to convert the known limit into the new ones.

Thank you in advance.

Hint for the third: The product of a function that goes to 0 with a bounded function must have a limit of 0.
 
I see. Can I use the same logic in the first as well ?
 
Yankel said:
I see. Can I use the same logic in the first as well ?

No, neither function goes to 0.

As for whether the function has an infinite limit, as it oscillates I would lean towards no...
 

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