Prove f'(x)=f(x)

Question

Given that "f" is a function with the following properties:

 

i. f(x₁+x₂)=f(x₁)f(x₂) for all x₁ and x₂

ii. f(x)=1+x*g(x) where lim_{x approaches 0} g(x)=1

 

Pover: f'(x)=f(x)

Answer 1

This one looks harder than it is. The whole proof is just the limit definition of the derivative, with each hypothesis used exactly once.

Start from the definition

f′(x) = lim_h→0 [ f(x + h) − f(x) ] / h

Use property (i) on f(x + h)

Property (i) says f(x₁ + x₂) = f(x₁)f(x₂). Taking x₁ = x and x₂ = h turns the sum in the numerator into a product:

f′(x) = lim_h→0 [ f(x)f(h) − f(x) ] / h

Now f(x) is a constant as far as this limit is concerned — h is the variable, not x — so factor it out:

f′(x) = f(x) · lim_h→0 [ f(h) − 1 ] / h

Look at what's happened. The x and the h have been completely separated. All that's left is to show that remaining limit equals 1.

Use property (ii) on the leftover limit

Property (ii) says f(x) = 1 + x·g(x). Written at h:

f(h) = 1 + h·g(h)  ⇒  f(h) − 1 = h·g(h)

So the quotient simplifies before you take any limit at all:

[ f(h) − 1 ] / h = h·g(h) / h = g(h)

And we're told lim_h→0 g(h) = 1. Therefore

f′(x) = f(x) · 1 = f(x)  ■

What each hypothesis was for

This is worth noticing, because it's the pattern behind most proofs of this type. Property (i) is what lets you pull f(x) out of the limit — it converts f(x + h) into f(x) times something that depends only on h. Property (ii) is what evaluates the piece left behind. Neither one does the job alone.

Notice also that (ii) quietly pins down f(0): setting x = 0 gives f(0) = 1 + 0·g(0) = 1. That's consistent with (i), since putting x₁ = x₂ = 0 gives f(0) = f(0)², forcing f(0) to be 0 or 1 — and (ii) picks out 1. If f(0) were 0, property (i) would make f identically zero.

What function is this?

You don't need it for the proof, but it helps to know what you've been describing. The only function satisfying f′ = f with f(0) = 1 is f(x) = e^x, and it does satisfy both hypotheses: e^x₁+x₂ = e^x₁e^x₂, and (e^x − 1)/x → 1 as x → 0.

So the exercise is really building e^x from two abstract properties rather than from a formula — which is exactly how the exponential function gets defined in a more careful treatment.

One small note: never write f′(x) = f(x) · lim[f(h) − 1]/h and then substitute h = 0 directly. At h = 0 that quotient is 0/0. The reason it works is that property (ii) lets you cancel the h algebraically first, leaving g(h), which does have a limit. Cancel first, take the limit second.