*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

**and**

*f(x)***are two**

*g(x)**continuous*functions defined for all real arguments

**AND**

*x***as**

*f(x)→L***AND**

*x→r***as**

*g(x)→M*

*x→L*THEN

**as**

*g(f(x))→M*

*x→r**Proof*

Since

**as**

*g(x)→M*

*x→L***∀**positive

**(however small)**

*ε***∃**

(|

*δ*:(|

*f(x)−L*| ≤*δ*) ⇒ |*g(f(x))−M*| ≤*ε*Since

**as**

*f(x)→L*

*x→r***∀**positive

**(however small)**

*δ***∃**

(|

*γ*:(|

*x−r*| ≤*γ*) ⇒ |*f(x)−L*| ≤*δ*So, for any chosen

**we can always find**

*ε***such that**

*γ***(|**

⇒ |

*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 of

**to include co-domain (set of all values) of**

*g(x)***.**

*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

**for all real**

*f(x) = 0*

*x***for all real**

*g(x) = 0***AND**

*x ≠ 0***for**

*g(x) = 1*

*x = 0*Now

**as**

*f(x)→L=0*

*x→r=0***as**

*g(x)→M=0*

*x→L=0*But

**as**

*g(f(x))→1≠M*

*x→r=0*3. Limit points participating in this theorem (

**) might be infinite as well. The proof will be very similar and can be left to self-study.**

*r, L, M**Examples*

**as**

*f(x)=1/x→0*

*x→+∞***as**

*g(x)=sin(x)/x→1*

*x→0*Therefore,

**as**

*sin(1/x)/(1/x)→1*

*x→+∞***as**

*f(x)=sin(x)/x→1*

*x→0***as**

*g(x)=2*^{x}→1

*x→0*Therefore,

**as**

*2*^{1−sin(x)/x}→1

*x→0*
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