## AmTran_Bus one year ago Help finding crazy limit!

1. AmTran_Bus

2. 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-.

3. 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

4. Michele_Laino

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

5. AmTran_Bus

Oh ok. Thanks to both of you.

6. AmTran_Bus

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

7. AmTran_Bus

My mistake, I miscalculated @Michele_Laino

8. 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 $$-\infty$$

9. 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

10. AmTran_Bus

ans was 1/4 :(

11. AmTran_Bus

Oh well, I can retake it.

12. anonymous

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

13. Michele_Laino

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

14. AmTran_Bus

It was like the attachment said.

15. Michele_Laino

it is a factor, as I can see

16. ganeshie8
17. 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|

18. Michele_Laino

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

19. 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.

20. Michele_Laino

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