Combination Theorem for Limits of Functions/Real

Theorem
Let $X$ be one of the standard number fields $\Q, \R, \C$.

Let $f$ and $g$ be functions defined on an open subset $S \subseteq X$, except possibly at the point $c \in S$.

Let $f$ and $g$ tend to the following limits:


 * $\displaystyle \lim_{x \to c} f \left({x}\right) = l, \lim_{x \to c} g \left({x}\right) = m$

Let $\lambda, \mu \in X$ be any point in $X$.

Then the following results hold:

Sum Rule

 * $\displaystyle \lim_{x \to c} \left({f \left({x}\right) + g \left({x}\right)}\right) = l + m$

Multiple Rule

 * $\displaystyle \lim_{x \to c} \left({\lambda f \left({x}\right)}\right) = \lambda l$

Combined Sum Rule

 * $\displaystyle \lim_{x \to c} \left({\lambda f \left({x}\right) + \mu g \left({x}\right)}\right) = \lambda l + \mu m$

Product Rule

 * $\displaystyle \lim_{x \to c} \left({f \left({x}\right) g \left({x}\right)}\right) = l m$

Quotient Rule

 * $\displaystyle \lim_{x \to c} \frac {f \left({x}\right)} {g \left({x}\right)} = \frac l m$, provided that $m \ne 0$. (In the case that $l = m = 0$, see L'Hôpital's Rule).

Continuity
If $f$ and $g$ are continuous on an open subset $S \subseteq X$, and $\lambda, \mu \in X$, then:


 * $f + g $ is continuous on $S$
 * $\lambda f$ is continuous on $S$
 * $\lambda f + \mu g $ is continuous on $S$
 * $f g$ is continuous on $S$
 * $\dfrac f g$ is continuous on $S$.

Proof
These results follow directly from the Combination Theorem for Sequences and Limit of Function by Convergent Sequences, as follows:

Let $\left \langle {x_n} \right \rangle$ be any sequence of points of $S$ such that $\forall n \in \N^*: x_n \ne c$ and $\displaystyle \lim_{n \to \infty} x_n = c$.

By Limit of Function by Convergent Sequences, $\displaystyle \lim_{n \to \infty} f \left({x_n}\right) = l$ and $\displaystyle \lim_{n \to \infty} g \left({x_n}\right) = m$.

By the Combination Theorem for Sequences:
 * $\displaystyle \lim_{n \to \infty} \left({f \left({x_n}\right) + g \left({x_n}\right)}\right) = l + m$
 * $\displaystyle \lim_{n \to \infty} \left({\lambda f \left({x_n}\right)}\right) = \lambda l$
 * $\displaystyle \lim_{n \to \infty} \left({\lambda f \left({x_n}\right) + \mu g \left({x_n}\right)}\right) = \lambda l + \mu m$
 * $\displaystyle \lim_{n \to \infty} \left({f \left({x_n}\right) g \left({x_n}\right)}\right) = l m$
 * $\displaystyle \lim_{n \to \infty} \frac {f \left({x_n}\right)} {g \left({x_n}\right)} = \frac l m$, provided that $m \ne 0$.

Applying Limit of Function by Convergent Sequences again, we get:


 * $\displaystyle \lim_{x \to c} \left({f \left({x}\right) + g \left({x}\right)}\right) = l + m$
 * $\displaystyle \lim_{x \to c} \left({\lambda f \left({x}\right)}\right) = \lambda l$
 * $\displaystyle \lim_{x \to c} \left({\lambda f \left({x}\right) + \mu g \left({x}\right)}\right) = \lambda l + \mu m$
 * $\displaystyle \lim_{x \to c} \left({f \left({x}\right) g \left({x}\right)}\right) = l m$
 * $\displaystyle \lim_{x \to c} \frac {f \left({x}\right)} {g \left({x}\right)} = \frac l m$, provided that $m \ne 0$.

Proof of Continuity
We have that:
 * The complex plane is a metric space
 * The real number line is a metric space
 * The rational numbers form a metric space.

Hence it is appropriate to use the defintion of continuity on a metric space.

It follows trivially from the definitions of continuity at a point and continuity on a metric space that, if $f$ and $g$ are continuous on $S$ that the assertions:


 * $f + g $ is continuous on $S$
 * $\lambda f$ is continuous on $S$
 * $\lambda f + \mu g$ is continuous on $S$
 * $f g$ is continuous on $S$
 * $\dfrac f g$ (where $g \ne 0$) is continuous on $S$

are true.