calculate limit, x -> 0, of (1 - cos(sin^5(x)) / [(e^(x^4) - 1)(sin(x^2) - x^2)]

[odelia]

I'm having troubles with calculating the following limit:

. . . . .\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, 0}\, \) \(\displaystyle \dfrac{1\, -\, \cos\left(\sin^5 (x)\right)}{\left(e^{x^4}\, -\, 1\right)\left(\sin(x^2)\, -\, x^2\right)}\)

Please elaborate if possible!

 

[Ishuda]

I'm having troubles with calculating the following limit:

. . . . .\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, 0}\, \) \(\displaystyle \dfrac{1\, -\, \cos\left(\sin^5 (x)\right)}{\left(e^{x^4}\, -\, 1\right)\left(\sin(x^2)\, -\, x^2\right)}\)

Please elaborate if possible! :smile:

What are your thoughts? What have you done so far? Please show us your work even if you feel that it is wrong so we may try to help you. You might also read
http://www.freemathhelp.com/forum/threads/78006-Read-Before-Posting

As in a lot of problems, there are several ways to approach the problem. l'Hospital's rule will probably eventually get you there. However, if allowed, I would go with the expanded series approach which goes something like: For x near zero,

sin(x) is about x-x³/6, cos(x) is about 1-x²/2, e^x is about 1+x. putting that with the given expression

1-cos(sin(x⁵)) ~ x¹⁰/2

\(\displaystyle e^{x^4}\, -\, 1\, \approx\, x^4\)

...

So, for x near zero

\(\displaystyle \dfrac{1\, -\, \cos\left(\sin^5(x)\right)}{\left(e^{x^4}\, -\, 1\right)\left(\sin(x^2)\, -\, x^2\right)}\, \approx\, \dfrac{x^{10}}{2\, x^4\, ...}\, ...\)

Of course you would need to be a bit more formal in the workings.