I am trying to find the following 2 limits as part of a series of 4 exercises following out lectures. We haven't really learned derivatives yet, so I can't just slap L'Hospital and be done with it.
The first two I could solve with Taylor Expansions, but I still need to solve:
$$\lim_{x\to 0} \frac{e^x -\sin x -1}{x^2}$$ Taylor Expansion of sin didn't work here, and I don't know how to deal with the $e^x$.
$$\lim_{x\to 0} \frac{x\sin x}{\cos x -1}$$ Given $-\sin^2 x = \cos^2 x -1$ I tried multiplying by $\frac{\cos x +1}{\cos x +1}$, which lead me to $\frac{-x\cos x \ -x}{\sin x}$ and then I don't know what else to do. (SOLVED) I suppose I could simplify this to $-x\cot x - \frac{x}{\sin x}$, which leads to -2.
You have
$$e^x=1+x+\frac{x^2}{2}(1+\epsilon_1(x))$$
and
$$\sin(x)=x+x^2\epsilon_2(x)$$
cause $x\mapsto \sin(x)$ is odd.
then the limit is $$\frac{1}{2}$$