I'm stuck with this problem:
If $f(x)=\frac{\sin x}{x}$ for $x\neq 0$ and $f(0)=1$, find $f^{(k)}(0)$. Hint: Find the power series of f.
I tried to solve it by writing down the Taylor series of $$\sin x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-...$$
After that I divided the whole series by $x$: $$\frac{\sin x}{x}=\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n+1)!}=f(x)$$. The problem is that when I plug in $0$ in $f(x)$, I get $f(0)=$undefined. Also, When I differentiate and plug in $0$ I always get $0$, what am I doing wrong? How can I solve this? Thanks.
Hint: If you know that $$f(x) = \sum\limits_{k=0}^{\infty}a_{k}x^{k},$$ then $$a_{k} = \dfrac{f^{(k)}(0)}{k!}.$$
(Try and show this!)
Also remember that if you are plugging in $x = 0$ to a term of a power series where the power of $x$ is $0$, the term should be $1$, not $0$ (i.e. substituting $x = 0$ into something like $x^{k}$ when $k = 0$ should be treated as $1$, not $0$).
For example, if $g(x) = \sum\limits_{k=0}^{\infty}\frac{1}{k!}x^{k}$, then if you substitute $x = 0$, what do you get?