Show that $\lim_{x\rightarrow0}\frac{e^{\iota x}-1}{\iota x}=1$

Question

(a) Show that $|e^{\iota a}-e^{\iota b}|\le\min\{2,|a-b|\}$ for all $a,b\in\mathbb{R}$. Hint. Write it as an integral from $a$ to $b$.
(b) Show that $$\lim_{x\rightarrow0}\frac{e^{\iota x}-1}{\iota x}=1$$

I have been able to prove part (a) by setting $f(x)=e^{\iota x}$ and using the expansion of $e^{\iota x}$. Assuming part (b) is not unrelated to part (a), how can I possibly prove it? I tried showing the LHS $\le1$ using the fact that $|e^{\iota a}-e^{\iota b}|\le|a-b|$, but am stuck with the other half. Any help?

Answer 1

$|e^{ix}-1| \leqslant \min (2,|x-0|) = |x|$, so if the limit is defined, it converges with module $\leq 1$. You can't have more from this. You can try working with the real part and the imaginary part but I don't know why you would bother.

Otherwise it is clearly the definition of $\frac{1}{i} \cdot\frac{d}{dx} (e^{ix})(0) = 1$.

Answer 2

Since this is $\dfrac{0}{0}$ we apply L'Hospital's rule \begin{align} f'(x) &= \frac{i \, exp(ix)}{i} \\ \lim_{x>0} f^{'}(x) &= e^{0} = 1 \end{align}