# limit rules of functions

## Primary tabs

Keywords:
limit rule
Synonym:
limit rules of sequences
Type of Math Object:
Theorem
Major Section:
Reference
Parent:
Groups audience:

## Mathematics Subject Classification

### limit rule of compound function

Hi activists,
The new member urscheljd has asked the proof of the Theorem 2 in http://planetmath.org/encyclopedia/LimitRulesOfFunctions.html
Is there someone member who would want to add such a proof in PM?
Regards,
Jussi

### limit rule of compound function

Hi activists,
The new member urscheljd has asked the proof of the Theorem 2 in http://planetmath.org/encyclopedia/LimitRulesOfFunctions.html
Is there someone member who would want to add such a proof in PM?
Regards,
Jussi

### Re: limit rule of compound function

I'm super busy with classes right now, but in the next week or so I might be able to add a proof, although I think it would be more beneficial to drop the assumption that g be continuous at $a$ and assume only that \lim_{x\rightarrow a}g(x) exists. In this case, we have

\lim_{x\rightarrow x_0}g(f(x))=\lim_{x\rightarrow a}g(x),

which implies the result as currently stated, since in this case

\lim_{x\rightarrow a}g(x)=g(a)=g(\lim_{x\rightarrow x_0}f(x)). Just a suggestion though.

### Re: limit rule of compound function

What you have isn't true in general, unfortunately. Consider a function $g(x)$ which isn't continuous at $a$ but that $\lim_{x\to a} g(x)$ exists. Then for $f(x)=a$, $\lim_{x\to x_0} g(f(x))=g(a)\neq \lim_{x\to a} g(x)$.

### Re: limit rule of compound function

Continuity of g at x=a is essential.
A\subset{R}---f--->B\subset{R}---g--->R(the reals)
x---------->y=f(x)---------->z=g(y)
A\subset{R}--------gof---------->R
x--------------------------->(gof)(x):=g(f(x))
codomain{f} \subset domain{g} \implies gof is defined.
\lim_{x\to x_0}f(x)=a (possibly not equals to f(x_0))
g *continuos* at y=a \implies \lim_{y\to a}g(y)=g(a),
so g(\lim_{x\to x_0}f(x))=\lim_{f(x)\to a} g(f(x)) =\lim_{x\to x_0}g(f(x)),
since as f(x)\to a then x\to x_0, as limit of f(x) exists at x=x_0 by hypothesis.

### Re: limit rule of compound function

It's possible I wasn't quite precise enough. Consider the following: let S, T be subsets of \mathbb{R}, f:S\rightarrow\mathbb{R}, g:T\rightarrow \mathbb{R}R, and f(S)\subset T. Fix a\in S, and assume \lim_{x\rightarrow a}f(x) exists and equals b. Assume further that \lim_{y\rightarrow b}g(y) exists and equals c. Then

\lim_{x\rightarrow a}g(f(x))=c.

Given \epsilon>0, there exists \delta_1>0 such that if y\in T and 0<|y-b|<\delta_1, then |g(y)-b|<\epsilon. Select \delta_2>0 such that for all x\in S with 0<|x-a|<\delta_1, |f(x)-b|<\delta_1. It follows that for all such x, |g(f(x))-b|<\epsilon, so that \lim_{x\rightarrow a}g(f(x))=b.