The Limit of (sinx+1)/(x) - 1 or Infinity?

[anemone]

Someone told me that the limit as x goes to zero of (sinx+1)/(x) is 1 but I think it's quite obvious that its limit is infinity. I hope you can tell me which is the correct answer? 1 or infinity? Many thanks in advance for replying.

 

[alexmahone]

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

 

[Plato2]

Someone told me that the limit as x goes to zero of (sinx+1)/(x) is 1 but I think it's quite obvious that its limit is infinity. I hope you can tell me which is the correct answer? 1 or infinity? Many thanks in advance for replying.

misread the OP

 

[CaptainBlack]

Someone told me that the limit as x goes to zero of (sinx+1)/(x) is 1 but I think it's quite obvious that its limit is infinity. I hope you can tell me which is the correct answer? 1 or infinity? Many thanks in advance for replying.

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

 

[CaptainBlack]

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

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

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

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

 

[alexmahone]

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

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

Oops...

 

[anemone]

Thanks to all!

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

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.:)

 

[melese]

Thanks 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.:)

To anemone,
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 different, and that means no limit at \(x=0\).

 

[CaptainBlack]

to anemone,
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 different, and that means no limit at \(x=0\).

The one-sided limits do not exist!

Cb