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- Feb 14, 2012

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- Jan 26, 2012

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$\displaystyle \lim_{x\to 0}\frac{\sin x+1}{x}=\infty$

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misread the OP

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- Jan 26, 2012

- 890

Are you sure that you have typed that correctly? Can you add brackets to remove any possible ambiguity?

But yes, as it stands the limit:

\[ \lim_{x \to 0} \frac{\sin(x)+1}{x}=\infty \]

but note:

\[ \lim_{x \to 0} \frac{\sin(x)}{x}=1 \]

CB

- Jan 26, 2012

- 890

Since \(\displaystyle \lim_{x\to 0} [\sin(x+1)]=\sin(1)\ne 0 \) your limit$\displaystyle \lim_{x\to 0}\frac{\sin x+1}{x}=\infty$

"Someone" may have been referring to $\displaystyle \lim_{x\to 0}\frac{\sin(x+1)}{x}$, which is 1.

\[ \lim_{x\to 0}\frac{\sin(x+1)}{x} =\lim_{x\to 0} \frac{\sin(1)}{x} =\infty \]

- Jan 26, 2012

- 268

Oops...Since \(\displaystyle \lim_{x\to 0} [\sin(x+1)]=\sin(1)\ne 0 \) your limit

\[ \lim_{x\to 0}\frac{\sin(x+1)}{x} = \infty \]

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- Feb 14, 2012

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Thanks to all!

\[ \lim_{x\to 0} \frac{sin(x)+1}{x} \]

Now you all have made it very clear that

\[ \lim_{x\to 0} \frac{sin(x)+1}{x}=\infty \]

That's my answer too!

Thanks.

I think I had better learn Latex if I want to ask more question(s) on this site.

OK and I'm sorry. I meant to ask:Are you sure that you have typed that correctly? Can you add brackets to remove any possible ambiguity?

\[ \lim_{x\to 0} \frac{sin(x)+1}{x} \]

Now you all have made it very clear that

\[ \lim_{x\to 0} \frac{sin(x)+1}{x}=\infty \]

That's my answer too!

Thanks.

I think I had better learn Latex if I want to ask more question(s) on this site.

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- Feb 24, 2012

- 27

ToThanks to all!

OK and I'm sorry. I meant to ask:

\[ \lim_{x\to 0} \frac{sin(x)+1}{x} \]

Now you all have made it very clear that

\[ \lim_{x\to 0} \frac{sin(x)+1}{x}=\infty \]

That's my answer too!

Thanks.

I think I had better learn Latex if I want to ask more question(s) on this site.

the limit \[\lim_{x\to 0} \frac{sin(x)+1}{x}\] does not exist! Reason: supose the limit existed, then \[ \lim_{x\to 0} \frac{sin(x)+1}{x}=\lim_{x\to 0}( \frac{sin(x)}{x}+\frac{1}{x})=\lim_{x\to 0} \frac{sin(x)}{x}+\lim_{x\to 0}\frac{1}{x}=1+\lim_{x\to 0}\frac{1}{x}\].

But, \(\displaystyle\lim_{x\to 0^+} \frac{1}{x}=\infty\), whereas \(\displaystyle\lim_{x\to 0^-} \frac{1}{x}=-\infty\); the point you should notice is the distinction of one-sided limits.

It follows that the two one-sided limits exits, but they're

- Jan 26, 2012

- 890

The one-sided limits do not exist!toanemone,

the limit \[\lim_{x\to 0} \frac{sin(x)+1}{x}\] does not exist! Reason: Supose the limit existed, then \[ \lim_{x\to 0} \frac{sin(x)+1}{x}=\lim_{x\to 0}( \frac{sin(x)}{x}+\frac{1}{x})=\lim_{x\to 0} \frac{sin(x)}{x}+\lim_{x\to 0}\frac{1}{x}=1+\lim_{x\to 0}\frac{1}{x}\].

But, \(\displaystyle\lim_{x\to 0^+} \frac{1}{x}=\infty\), whereas \(\displaystyle\lim_{x\to 0^-} \frac{1}{x}=-\infty\); the point you should notice is the distinction of one-sided limits.

It follows that the two one-sided limits exits, but they'redifferent, and that means no limit at \(x=0\).

Cb