AmTran_Bus
  • AmTran_Bus
Help finding crazy limit!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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AmTran_Bus
  • AmTran_Bus
AmTran_Bus
  • AmTran_Bus
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-.
ganeshie8
  • ganeshie8
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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Michele_Laino
  • Michele_Laino
hint: we can write this: \[\Large \frac{{{e^x} - 6}}{{\sin \left( {4x} \right)}} = \frac{{{e^x} - 6}}{{4x}}\frac{1}{{\frac{{\sin \left( {4x} \right)}}{{\left( {4x} \right)}}}}\]
AmTran_Bus
  • AmTran_Bus
Oh ok. Thanks to both of you.
AmTran_Bus
  • AmTran_Bus
Wait, isnt that going to come out as one @Michele_Laino And ganeshie8 said -inf?
AmTran_Bus
  • AmTran_Bus
My mistake, I miscalculated @Michele_Laino
anonymous
  • anonymous
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 \(-\infty\)
ganeshie8
  • ganeshie8
it is -infy if you let sin(4x) be sin^4(x) but as the expression stands, the limit is DNE as you said
AmTran_Bus
  • AmTran_Bus
ans was 1/4 :(
AmTran_Bus
  • AmTran_Bus
Oh well, I can retake it.
anonymous
  • anonymous
Oops, didnt recognize ganeshie's wolfram link was for sin^4(x), lol
Michele_Laino
  • Michele_Laino
I don't understand: 4 is an exponent or it is a factor?
AmTran_Bus
  • AmTran_Bus
It was like the attachment said.
Michele_Laino
  • Michele_Laino
it is a factor, as I can see
ganeshie8
  • ganeshie8
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
Michele_Laino
  • Michele_Laino
in that case, your limit doesn't exist since we have to consider the graph of the fuinction 1/x, namely: |dw:1433615875097:dw|
Michele_Laino
  • Michele_Laino
I think we have this: \[\large \begin{gathered} \mathop {\lim }\limits_{x \to 0 + } \frac{{{e^x} - 6}}{{\sin \left( {4x} \right)}} = \mathop {\lim }\limits_{x \to 0 + } \frac{{{e^x} - 6}}{{4x}} \times \mathop {\lim }\limits_{x \to 0 + } \frac{1}{{\frac{{\sin \left( {4x} \right)}}{{\left( {4x} \right)}}}} = - \infty \hfill \\ \hfill \\ \mathop {\lim }\limits_{x \to 0 - } \frac{{{e^x} - 6}}{{\sin \left( {4x} \right)}} = \mathop {\lim }\limits_{x \to 0 - } \frac{{{e^x} - 6}}{{4x}} \times \mathop {\lim }\limits_{x \to 0 - } \frac{1}{{\frac{{\sin \left( {4x} \right)}}{{\left( {4x} \right)}}}} = + \infty \hfill \\ \end{gathered} \]
anonymous
  • anonymous
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.
Michele_Laino
  • Michele_Laino
please substitute x=0.01, into your original expression, what do you get?

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