TITLE: [Stuck]: Final step of solving limit, Calculus I
QUESTION [5 upvotes]: Calculate the value of $k$ such that the following limit has a finite solution, $L$ such that $L \ne 0$: 
$$\lim_{x\rightarrow0} \frac{(e^{x^2}-x^2-1)(\cos(x)-1)}{x^k}$$
I use the Taylor Series expansions of $e^x$ and $\cos(x)$ and simplify the above expression to the following:
$$\lim_{x\rightarrow0} \frac{-\frac{1}{4}x^6+(\frac{1}{48}-\frac{1}{12})x^8+(\frac{1}{144})x^{10}}{x^k}$$ 
Now I have a mental roadblock. For $k<6$ the above limit goes to zero and for $k>6$ this expression should diverge but in my head it goes to zero again.... I am trying to think of a simple example to convince myself but can't. Can someone please help me understand this? 
Thanks.

REPLY [2 votes]: As an alternative derivation, we have that
$$\lim_{x\rightarrow0} \frac{(e^{x^2}-x^2-1)(\cos x-1)}{x^k}
=\lim_{x\rightarrow0} \frac{e^{x^2}-x^2-1}{x^{k-2}}\lim_{x\rightarrow0} \frac{\cos x-1}{x^2}=-\frac12\lim_{x\rightarrow0} \frac{e^{x^2}-x^2-1}{x^{k-2}}$$
then recall that 
$$e^{x^2}=1+x^2+\frac12x^4+o(x^4)$$
and then for $k=6$
$$\lim_{x\rightarrow0} \frac{e^{x^2}-x^2-1}{x^{4}}=\lim_{x\rightarrow0} \frac{\frac12x^4+o(x^4)}{x^{4}}=\lim_{x\rightarrow0} \frac{\frac12+o(1)}{1}=\frac12$$
and therefore
$$\lim_{x\rightarrow0} \frac{(e^{x^2}-x^2-1)(\cos x-1)}{x^6}=-\frac14$$