Question

Help finding crazy limit!

Answer 1

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-.

Answer 2

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

Answer 3

hint:
we can write this:
\[\Large \frac{{{e^x} - 6}}{{\sin \left( {4x} \right)}} = \frac{{{e^x} - 6}}{{4x}}\frac{1}{{\frac{{\sin \left( {4x} \right)}}{{\left( {4x} \right)}}}}\]

Answer 4

Oh ok. Thanks to both of you.

Answer 5

Wait, isnt that going to come out as one @Michele_Laino
And ganeshie8 said -inf?

Answer 6

My mistake, I miscalculated @Michele_Laino

Answer 7

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 \(-\infty\)

Answer 8

it is -infy if you let sin(4x) be sin^4(x)

but as the expression stands, the limit is DNE as you said

Answer 9

ans was 1/4 :(

Answer 10

Oh well, I can retake it.

Answer 11

Oops, didnt recognize ganeshie's wolfram link was for sin^4(x), lol

Answer 12

I don't understand:
4 is an exponent or it is a factor?

Answer 13

It was like the attachment said.

Answer 14

it is a factor, as I can see

Answer 15

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

Answer 16

in that case, your limit doesn't exist since we have to consider the graph of the fuinction 1/x, namely:

Answer 17

I think we have this:
\[\large \begin{gathered}
\mathop {\lim }\limits_{x \to 0 + } \frac{{{e^x} - 6}}{{\sin \left( {4x} \right)}} = \mathop {\lim }\limits_{x \to 0 + } \frac{{{e^x} - 6}}{{4x}} \times \mathop {\lim }\limits_{x \to 0 + } \frac{1}{{\frac{{\sin \left( {4x} \right)}}{{\left( {4x} \right)}}}} = - \infty \hfill \\
\hfill \\
\mathop {\lim }\limits_{x \to 0 - } \frac{{{e^x} - 6}}{{\sin \left( {4x} \right)}} = \mathop {\lim }\limits_{x \to 0 - } \frac{{{e^x} - 6}}{{4x}} \times \mathop {\lim }\limits_{x \to 0 - } \frac{1}{{\frac{{\sin \left( {4x} \right)}}{{\left( {4x} \right)}}}} = + \infty \hfill \\
\end{gathered} \]

Answer 18

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.

Answer 19

please substitute x=0.01, into your original expression, what do you get?