# Show that $\lim_{h\rightarrow 0}\frac{e^{ih}-1}{h}=i$

$$h\in \mathbb{R}$$, because we have defined the Trigonometric Functions only on $$\mathbb{R}$$ so far.

I have a look at $$e^{ih}=\sum_{k=0}^{\infty}\frac{(ih)^k}{k!}=1+ih-\frac{h^2}{2}+....$$

How can one describe the nth term of the sum?

Then I look at $$\frac{e^{ih}-1}{h}=\frac{(1-1)}{h}+i-\frac{h}{2}+...=i-\frac{h}{2}+....$$

Again how can I describe that the nth term of the sum?

Because $$\frac{e^{ih}-1}{h}=\sum_{k=1}^{\infty}\frac{\frac{(ih)^k}{h}}{k!}<\sum_{k=0}^{\infty}\frac{\frac{(ih)^k}{h}}{k!}=\sum_{k=1}^{\infty}\frac{(ih^{-1})^k}{k!}=e^{ih^{-1}}$$

and $$ih^{-1}$$ is a complex number and the exponential-series converges absolutely for all Elements in $$\mathbb{C}$$, I have found a convergent majorant. And I can apply the properties of Limits on $$\frac{e^{ih}-1}{h}\forall, h\in \mathbb{R}$$.

How can I now prove formally (i.e by chosing an explicit $$\delta$$) that

$$\forall_{\epsilon>0}\exists_{\delta>0}\forall_{h\in\mathbb{R}}|h-0|=|h|<\delta\Longrightarrow |(\frac{e^{ih}-1}{h}=i-\frac{h}{2}+...)-i|<\epsilon$$

I am also seeking for advice how to argue in such cases more intuitively (i.e by not always ginving an explicit $$\delta$$ ).

• (1) for the very first one, you already have an expression for the $k$th term, otherwise how can you use sigma notation for a sum? (2) for the bit before you ask for a formal proof, you can't use inequalities when you are trying to compare complex numbers in $\mathbb C$ – Calvin Khor Jan 23 at 13:42
• @CalvinKhor On (2), If I would put $|\cdot |$ around the sums would that work? – RM777 Jan 23 at 13:47
• You should check absolute convergence, because this is something that extends to complex numbers nicely (and has $|\cdot|$s everywhere) – Calvin Khor Jan 23 at 13:49

Sketch that follows the spirit of your approach, rather than using trigonometry etc. A useful technical tool:

Theorem. Suppose the continuous functions $$f_n=f_n(h)$$ taking values in $$\mathbb C$$ are such that $$\sum_{n=0}^\infty f_n(h) := \lim_{N\to\infty} \sum_{n=0}^N f_n(h)$$ converges absolutely uniformly on some interval $$h\in [-a,a]$$ (i.e. $$\sum_{n=0}^\infty \|f_n\|_{\infty} < \infty$$). Then $$\lim_{h\to 0} \sum_{n=0}^\infty f_n(h) = \sum_{n=0}^\infty f_n(0)$$

Sketch proof of theorem: Recall that the uniform limit of continuous functions is continuous. (see Proof of uniform limit of Continuous Functions or Wikipedia ). The $$N$$th partial sums $$F_N(h) := \sum_{n=0}^N f_n(h)$$ are continuous, and they converge pointwise on $$[-a,a]$$ to $$F(h):=\sum_{n=0}^\infty f_n(h)$$, and the absolutely uniformly convergent assumption implies that this convergence is uniform on $$[-a,a]$$. Hence, $$F$$ is continuous at all points in $$[-a,a]$$, and in particular at $$0$$. The result follows.

Application: $$\frac{e^{ih}-1}{h}\\= \frac{(\sum_{n=0}^\infty (ih)^n/n! )- 1} {h} \\= \frac{\sum_{n=1}^\infty (ih)^n/n! } {h} \\= \sum_{n=1}^\infty \frac{i(ih)^{n-1}}{n!} \\= \sum_{m=0}^\infty \frac{i(ih)^{m}}{(m+1)!}$$ After you verify we can apply the theorem, this yields $$\lim_{h\to 0 } \frac{e^{ih}-1}{h} = i+0+0+\dots = i$$.

PS make sure you realise that all infinite sums are defined in terms of limits in $$\mathbb C$$ (e.g. $$a_n \to a \in\mathbb C$$ iff $$|a_n - a| \to 0$$)

PPS this works to find the derivative at an arbitrary point of any function expressed as (for example) a Taylor series.

• Nice answer, Calvin. I was wondering if you had a source or proof of the theorem? – Jam Jan 23 at 14:15
• @Jam it essentially follows from the fact that the uniform limit of continuous functions is continuous (and the sequence of partial sums is a sequence of continuous functions that converge uniformly under the above assumptions). Does this help? – Calvin Khor Jan 23 at 14:16
• It does. Thank you – Jam Jan 23 at 14:17
• @Jam you're welcome – Calvin Khor Jan 23 at 14:21
• @RM777 I have added the proof to the answer, as in the comment to Jam – Calvin Khor Jan 23 at 22:11

Hint: Use Euler's formula and split the limit into well known trigonometric limits.

Here is a simple proof which I first read in Hardy's A Course of Pure Mathematics.

To reiterate the question for clarity we have $$h\in\mathbb{R}$$ and the symbol $$e^{ih}$$ is defined by the series $$\sum_{n=0}^{\infty} (ih) ^n/n!$$. Let's assume $$|h|< 1$$ and observe that \begin{align*} \left|\frac {e^{ih} - 1}{h}-i\right|&=\left|\frac{i^2h}{2!}+\frac{i^3h^2}{3!}+\dots\right|\\ &\leq\frac{|h|}{2!}+\frac{|h|^2}{3!}+\dots\\ &\leq\frac{|h|} {2}+\frac{|h|^2}{2^2}+\frac{|h|^3}{2^3}+\dots\\ &=\frac{|h|}{2-|h|}\text{ (sum of infinite GP)} \\ &<|h| \end{align*} Thus given any $$\epsilon>0$$ if we choose $$\delta=\min(1,\epsilon)$$ then we have $$\left|\frac{e^{ih} - 1}{h}-i\right|<\epsilon$$ whenever $$0<|h|<\delta$$. Therefore by definition of limit we have $$\lim_{h\to 0}\dfrac{e^{ih}-1}{h}=i$$

Hardy uses the above limit to prove the formula $$e^{ix} =\cos x+i\sin x \,\forall x\in\mathbb {R}$$ The idea is to show that the function $$f(x) =e^{ix}$$ is differentiable with derivative $$f'(x) =if(x)$$ and then consider the function $$g(x) =(\cos x-i\sin x) f(x)$$ It is easily proved that $$g'(x) =0$$ and hence $$g$$ is constant. Thus $$g(x) =g(0)=1$$ and $$f(x) =\cos x+i\sin x$$.

Show that $$\lim_{h\rightarrow 0}\frac{e^{ih}-1}{h}=i$$:

As we know: $$\cos \theta + i \sin \theta = e^{i \theta}$$

So, $$\lim \limits_{h\rightarrow 0} [\frac { \cos h + i \sin h - 1 } {h} ] = \lim \limits_{h \rightarrow 0} [ \frac {1 - \frac{h^2}{2!} + \phi_1 (h^4) + i( h - \frac{h^3}{3!}... \phi_2(h^5) )-1} {h} ] = \lim \limits_{h \rightarrow0} \frac { i(1 - higher \space powers\space of\space h) }{1} = i$$

Alternatively,

$$e^{ih} = 1 + ih + \frac{-h^2}{2!}...$$ will give the same result, just take terms of $$i$$ to one side and none $$i$$ ones to other, then take $$i$$ common and cancel 1 from numerator, then take $$h$$ common and cancel it from denominator, you'll get the desired result!