For sin x ln x, I'm thinking...
\sin x \ln x \sim x \ln x
And
\lim_{x\rightarrow0}x \ln x = \lim_{x\rightarrow\infty} \frac{1}{x}\ln\frac{1}{x} = \lim_{x\rightarrow\infty} -\frac{\ln x}{x} = 0
Therefore:
\lim_{x\rightarrow0}(5+\sin x \ln x) = 5
For the first lot, do I find Taylor...
I'm sorry that's what I meant. If you look at the limit I'm looking for, it's e^{x^2} - 1 \sim x^2 therefore sin^3 x(e^{x^2}-1) \sim x^5.
Now what about (x-\sinh x)(\cosh x- \cos x) and 5+\sin x \ln x?
Thanks!
Why do the others terms (9x2/2 + 9x3/2...) disappear as x->0?
PS: I had thought of replacing \sin^3 x (e^{x^2}) \sim x^5 because I know that \sin [f(x)] \sim f(x) and e^{f(x)} -1 \sim f(x). The rest I have no clue though.
Homework Statement
I need to find the following limit.
Homework Equations
\lim_{x\rightarrow0}\frac{(x-\sinh x)(\cosh x- \cos x)}{(5+\sin x \ln x) \sin^3 x (e^{x^2}-1)}
The Attempt at a Solution
I think it's got to be something with Taylor series, but I don't really know how to do it.