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

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The limit of (sin(x) + 1)/x as x approaches zero is debated, with some asserting it equals infinity while others argue it does not exist. The correct interpretation shows that the limit diverges due to the behavior of the term 1/x, which approaches infinity from the right and negative infinity from the left. Thus, the limit does not exist because the one-sided limits differ. Clarifications were made regarding potential misinterpretations of the expression, emphasizing the importance of precise notation. Overall, the consensus is that the limit does not exist at x = 0.
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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.
 
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$\displaystyle \lim_{x\to 0}\frac{\sin x+1}{x}=\infty$
 
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anemone said:
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
 
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anemone said:
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
 
Alexmahone said:
$\displaystyle \lim_{x\to 0}\frac{\sin x+1}{x}=\infty$

"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

\[ \lim_{x\to 0}\frac{\sin(x+1)}{x} =\lim_{x\to 0} \frac{\sin(1)}{x} =\infty \]
 
CaptainBlack said:
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...
 
Thanks to all!

CaptainBlack said:
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.:)
 
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anemone said:
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 \(x=0\).
 
melese said:
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
 
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