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Homelimit rules of functions

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# limit rules of functions

###### Theorem 1.

Let $f$ and $g$ be two real or complex functions. Suppose that there exist the limits $\lim_{{x\to x_{0}}}f(x)$ and $\lim_{{x\to x_{0}}}g(x)$. Then there exist the limits $\lim_{{x\to x_{0}}}[f(x)\!\pm\!g(x)]$, $\lim_{{x\to x_{0}}}f(x)g(x)$ and, if $\lim_{{x\to x_{0}}}g(x)\neq 0$, also $\lim_{{x\to x_{0}}}f(x)/g(x)$, and

1. $\lim_{{x\to x_{0}}}[f(x)\!\pm\!g(x)]\;=\;\lim_{{x\to x_{0}}}f(x)\pm\lim_{{x\to x% _{0}}}g(x),$

2. $\lim_{{x\to x_{0}}}f(x)g(x)\;=\;\lim_{{x\to x_{0}}}f(x)\cdot\lim_{{x\to x_{0}}% }g(x),$

3. $\lim_{{x\to x_{0}}}\frac{f(x)}{g(x)}\;=\;\frac{\lim_{{x\to x_{0}}}f(x)}{\lim_{% {x\to x_{0}}}g(x)},$

4. $\lim_{{x\to x_{0}}}c\;=\;c\quad\mathrm{where}\,\,c\,\,\mathrm{is\,\,a\,\,% constant}.$

These rules are used in limit calculations and in proving the corresponding differentiation rules (sum rule, product rule etc.).

In theorem 1, the domains of $f$ and $g$ could be any topological space (not necessarily $\mathbb{R}$ or $\mathbb{C}$).

As well, one often needs the

###### Theorem 2.

If there exists the limit $\lim_{{x\to x_{0}}}f(x)=a$ and if $g$ is continuous at the point $x=a$, then there exists the limit $\lim_{{x\to x_{0}}}g(f(x))$, and

$\lim_{{x\to x_{0}}}g(f(x))\;=\;g(\lim_{{x\to x_{0}}}f(x)).$ |

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26A06*no label found*30A99

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## Comments

## 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.