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So, I'm given the following query:

Write the Taylor series centered at $z_0 = 0$ for each of the following complex-valued functions:

$$f(z) = z^2\sin(z),\quad g(z) = z\sin(z^2)$$

Then, use these series to help you compute the following limit:

$$\lim_{z \to 0} \frac{z^2\sin(z)-z\sin(z^2)}{z^5}$$

So, the first part wasn't so bad. I simply noticed that

$$\sin(z) = \sum_{n=0}^{\infty} \frac{(-1)^nz^{2n+1}}{(2n+1)!}$$ and $$\sin(z^2) = \sum_{n=0}^{\infty} \frac{(-1)^nz^{4n+2}}{(2n+1)!}$$

With a little bit of simplifying, I obtained:

$$f(z) = \sum_{n=0}^{\infty} \frac{(-1)^nz^{2n+3}}{(2n+1)!}, \quad g(z) = \sum_{n=0}^{\infty} \frac{(-1)^nz^{4n+3}}{(2n+1)!}$$

However, I'm not quite sure how to deal with the limit. Does anyone have any advice on how I could approach it? I've tried expressing the entire limit as a series, but this doesn't seem to simplify much, and the limit cares about what happens as $z \to 0$ instead of as $n \to \infty$ (as we usually think about with series).

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\begin{align} \lim_{z\to0} \frac{z^2\sin(z)-z\sin(z^2)}{z^5} &= \lim_{z\to0} \frac{\sum_{n=0}^{\infty} \dfrac{(-1)^nz^{2n+3}}{(2n+1)!}- \sum_{n=0}^{\infty} \dfrac{(-1)^nz^{4n+3}}{(2n+1)!}}{z^5} \\ &= \lim_{z\to0} \frac{\left(z^3-\dfrac{z^5}{3!}+\dfrac{z^7}{5!}-\cdots\right)-\left(z^3-\dfrac{z^7}{3!}+\dfrac{z^{11}}{5!}-\cdots\right)}{z^5} \\ &= \lim_{z\to0} -\dfrac{1}{3!}+\left(\dfrac{1}{5!}+\dfrac{1}{3!}\right)z^7+P(z) \\ &= -\dfrac{1}{3!} \end{align} where $P(z)$ is an analytic function with $a_0=0$.

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Note that for every power series $P(z)$ with constant term $c$ we have $\lim_{z \to 0} P(z) = c$. So the question is: in the entire limit represented as series, do we have a constant term? Do we have a summand with $z^n$ for some negative $n$?

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