Unizor - Derivatives - Limit of Compound Functions

Notes to a video lecture on http://www.unizor.com

Function Limit - Compounding

The following theorem might be very useful in solving the problems with limits;

Theorem

IF
f(x) and g(x) are twocontinuous functions defined for all real arguments x AND
f(x)→L as x→r AND
g(x)→M as x→L
THEN
g(f(x))→M as x→r

Proof

Since g(x)→M as x→L
∀ positive ε (however small)
∃δ:
(|f(x)−L| ≤ δ) ⇒ |g(f(x))−M| ≤ ε

Since f(x)→L as x→r
∀ positive δ (however small)
∃γ:
(|x−r| ≤ γ) ⇒ |f(x)−L| ≤ δ

So, for any chosen ε we can always find γ such that
(|x−r| ≤ γ) ⇒ |f(x)−L| ≤ δ ⇒
⇒ |g(f(x))−M| ≤ ε
End of proof.

Notes:

1. Condition of the domain of each of these functions to be all real numbers can be reduced to a condition of the domain ofg(x) to include co-domain (set of all values) of f(x).

2. Condition of continuity of both functions is mandatory for this theorem. The proof relies on it.
An example where continuity is important is as follows
f(x) = 0 for all real x
g(x) = 0 for all real x ≠ 0 AND
g(x) = 1 for x = 0
Now
f(x)→L=0 as x→r=0
g(x)→M=0 as x→L=0
But
g(f(x))→1≠M as x→r=0

3. Limit points participating in this theorem (r, L, M) might be infinite as well. The proof will be very similar and can be left to self-study.

Examples

f(x)=1/x→0 as x→+∞
g(x)=sin(x)/x→1 as x→0
Therefore,
sin(1/x)/(1/x)→1 as x→+∞

f(x)=sin(x)/x→1 as x→0
g(x)=2^x→1 as x→0
Therefore,
2^1−sin(x)/x→1 as x→0