2. Compute $\lim _{x \rightarrow 0} \frac{e^{x \cos x}-1-x}{\sin \left(x^{2}\right)}$.
Answer: $\frac{1}{2}$
Solution: Let's compute all the relevant Maclaurin series expansions, up to the quadratic terms:
$$
x \cos x=x+\ldots, \quad e^{x \cos x}=1+x+\frac{1}{2} x^{2}+\ldots, \quad \sin \left(x^{2}\right)=x^{2}+\ldots,
$$
so
$$
\lim _{x \rightarrow 0} \frac{e^{x \cos x}-1-x}{\sin \left(x^{2}\right)}=\lim _{x \rightarrow 0} \frac{\frac{1}{2} x^{2}+\ldots}{x^{2}+\ldots}=\frac{1}{2} .
$$