Limit of Function, Taylor Series

Question

I want to evaluate $$\lim_{t\to 0} \frac{e^{it}-1-zt}{t^2}$$ By expanding in Taylor Series and doing the permitted algebraic simplifications (mainly subtracting the $1$ from the taylor expansion of $\cos(t)$ and then canceling one t from the denominator), I obtain $$ \lim_{t\to 0} \frac{(\frac{-t}{2!}+\frac{t^3}{4!} +)... + i(1-\frac{t^2}{3!}-\frac{t^4}{5!}+...)-z}{t}$$ I'd argue that now I'm left with $$\lim_{t\to0}\frac{i-z}{t}$$ which goes to $\infty$. Can you confirm or correct my reasoning?

Answer 1

Put $z=a+ib$. then compute

$$\lim_{t\to 0}\frac{\cos(t)-1-at+i(\sin(t)-bt)}{t^2}$$

$$=\lim_{t\to 0}(\frac{-\frac{t^2}{2}-at+it(1-b)}{t^2}+\epsilon(t))$$

• If $b=1$ and $a=0$ then the limit is $\frac{-1}{2}.$ else, it doesn't exist.