Help finding crazy limit!

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Help finding crazy limit!

Mathematics
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I need to find the limit as x approaches zero of (e^x -6)/(sin 4x) The problem is the online homework wants the limit as x approaches 0, but don t only the left and right hand limits exist? Check the atached image for the solution choices they provide. I thought it was negative infinity for lim as x approaches 0+ and infinity for x approaches 0-.
You're correct. Looks there is a typographical error, the exponent 4 should be in the exponent i guess http://www.wolframalpha.com/input/?i=lim%28x%5Cto+0%29%5Cfrac%7B%5Cleft%28e%5Ex-6%5Cright%29%7D%7B%5Csin%5E4+%5Cleft%28x%5Cright%29%7D

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hint: we can write this: \[\Large \frac{{{e^x} - 6}}{{\sin \left( {4x} \right)}} = \frac{{{e^x} - 6}}{{4x}}\frac{1}{{\frac{{\sin \left( {4x} \right)}}{{\left( {4x} \right)}}}}\]
Oh ok. Thanks to both of you.
Wait, isnt that going to come out as one @Michele_Laino And ganeshie8 said -inf?
My mistake, I miscalculated @Michele_Laino
Doing what Michele did helps you get rid of the sin(4x) portion of it. Which just makes it easier to calculate the left and right hand limits to show that it is \(-\infty\)
it is -infy if you let sin(4x) be sin^4(x) but as the expression stands, the limit is DNE as you said
ans was 1/4 :(
Oh well, I can retake it.
Oops, didnt recognize ganeshie's wolfram link was for sin^4(x), lol
I don't understand: 4 is an exponent or it is a factor?
It was like the attachment said.
it is a factor, as I can see
http://www.wolframalpha.com/input/?i=lim%28x%5Cto+0%29%5Cfrac%7B%5Cleft%28e%5Ex-1%5Cright%29%7D%7B%5Csin+%5Cleft%284x%5Cright%29%7D
in that case, your limit doesn't exist since we have to consider the graph of the fuinction 1/x, namely: |dw:1433615875097:dw|
I think we have this: \[\large \begin{gathered} \mathop {\lim }\limits_{x \to 0 + } \frac{{{e^x} - 6}}{{\sin \left( {4x} \right)}} = \mathop {\lim }\limits_{x \to 0 + } \frac{{{e^x} - 6}}{{4x}} \times \mathop {\lim }\limits_{x \to 0 + } \frac{1}{{\frac{{\sin \left( {4x} \right)}}{{\left( {4x} \right)}}}} = - \infty \hfill \\ \hfill \\ \mathop {\lim }\limits_{x \to 0 - } \frac{{{e^x} - 6}}{{\sin \left( {4x} \right)}} = \mathop {\lim }\limits_{x \to 0 - } \frac{{{e^x} - 6}}{{4x}} \times \mathop {\lim }\limits_{x \to 0 - } \frac{1}{{\frac{{\sin \left( {4x} \right)}}{{\left( {4x} \right)}}}} = + \infty \hfill \\ \end{gathered} \]
Yeah, that is what you have. Something wrong with the question. You would think it was meant to be one of the variations ganeshie posted.
please substitute x=0.01, into your original expression, what do you get?

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