Properties of the limit of functions

The limit of functions has properties that are similar to the limit of sequences.

Theorem (Properties of the limit of function)

Let $f(x)$ and $g(x)$ be functions such that $\underset{x\to a}{lim}f(x)=\alpha$ and $\underset{x\to a}{lim}g(x)=\beta$. The following hold.

1. For any constants $k,l\in \mathbb{R}$, $$\underset{x\to a}{lim}(kf(x)+lg(x))=k\alpha +l\beta .$$
2. $$\underset{x\to a}{lim}f(x)g(x)=\alpha \beta .$$
3. If $\beta \ne 0$, $$\underset{x\to a}{lim}\frac{f(x)}{g(x)}=\frac{\alpha }{\beta }$$

Proof. In the following $\epsilon$ is always an arbitrary positive real number, and $D_f$ and $D_g$ are the domains of the functions $f$ and $g$, respectively.

1. Let $M=max\{|k|,|l|\}+1$. Then $\frac{\epsilon }{2M}$ is also positive and real. Since $\underset{x\to a}{lim}f(x)=\alpha$, there exists ${\delta }_1>0$ such that for any $x\in D_f$ if $0<|x-a|<{\delta }_1$ then $|f(x)-\alpha |<\frac{\epsilon }{2M}$. Similarly, there exists ${\delta }_2>0$ such that for any $x\in D_g$ if $0<|x-a|<{\delta }_2$ then $|g(x)-\beta |<\frac{\epsilon }{2M}$. Let $\delta =min\{{\delta }_1,{\delta }_2\}$. If $|x-a|<\delta$, then $$\begin{array}{rcl}|(kf(x)+lg(x))-(k\alpha +l\beta )|& =& |k(f(x)-\alpha )+l(g(x)-\beta )|\\ & \le & |k||f(x)-\alpha |+|l||g(x)-\beta |\\ & <& M\cdot \frac{\epsilon }{2M}+M\cdot \frac{\epsilon }{2M}\\ & =& \epsilon .\end{array}$$ Hence, $$\underset{x\to a}{lim}(kf(x)+lg(x))=k\alpha +l\beta .$$

2. Since $\underset{x\to a}{lim}f(x)=\alpha$, there exists ${\delta }_0>0$ such that for any $x\in D_f$, $0<|x-a|<{\delta }_0$ implies $|f(x)-\alpha |<\epsilon$. Then $\alpha -\epsilon <f(x)<\alpha +\epsilon$, and hence

$$|f(x)|\le max\{|\alpha -\epsilon |,|\alpha +\epsilon |\}.$$ Let us define $$M=max\{|\alpha -\epsilon |,|\alpha +\epsilon |,|\beta |\}.$$ Since $\epsilon >0$, at least one of $|\alpha -\epsilon |$ or $|\alpha +\epsilon |$ is not 0 so that $M>0$. Again, we can find ${\delta }_1>0$ such that $|x-a|<{\delta }_1$ implies $|f(x)-\alpha |<\frac{\epsilon }{2M}$. Similarly, we can find ${\delta }_2>0$ such that $|x-a|<{\delta }_2$ implies $|g(x)-\beta |<\frac{\epsilon }{2M}$. Let $\delta =min\{{\delta }_0,{\delta }_1,{\delta }_2\}$. If $0<|x-a|<\delta$, we have $$\begin{array}{rcl}|f(x)g(x)-\alpha \beta |& =& |f(x)g(x)-f(x)\beta +f(x)\beta -\alpha \beta |\\ & \le & |f(x)g(x)-f(x)\beta |+|f(x)\beta -\alpha \beta |\\ & =& |f(x)||g(x)-\beta |+|f(x)-\alpha ||\beta |\\ & <& M\cdot \frac{\epsilon }{2M}+\frac{\epsilon }{2M}\cdot M\\ & =& \epsilon .\end{array}$$ Hence, $$\underset{x\to a}{lim}f(x)g(x)=\alpha \beta .$$

3. It suffices to show that $$\underset{x\to a}{lim}\frac{1}{g(x)}=\frac{1}{\beta }.$$ We can find ${\delta }_0>0$ such that $0<|x-a|<{\delta }_0$ implies $|g(x)-\beta |<\frac{|\beta |}{2}$. By the triangle inequality, $$||g(x)|-|\beta ||\le |g(x)-\beta |<\frac{|\beta |}{2},$$ we have, in particular, $$|g(x)|>\frac{|\beta |}{2}>0$$ so that $$\frac{1}{|g(x)|}<\frac{2}{|\beta |}.$$ Again, we can find ${\delta }_1>0$ such that $|x-a|<{\delta }_1$ implies $|g(x)-\beta |<\frac{|\beta {|}^2\epsilon }{2}$. Let $\delta =min\{{\delta }_0,{\delta }_1\}$. If $0<|x-a|<\delta$, we have $$\begin{array}{rcl}|\frac{1}{g(x)}-\frac{1}{\beta }|& =& \left|\frac{\beta -g(x)}{g(x)\beta }\right|\\ & =& \frac{|g(x)-\beta |}{|g(x)||\beta |}\\ & <& \frac{|\beta {|}^2\epsilon }{2}\cdot \frac{2}{|\beta |}\cdot \frac{1}{|\beta |}\\ & =& \epsilon .\end{array}$$

■

Remark. Part 1 of this theorem shows that

$$\underset{x\to a}{lim}(kf(x)+lg(x))=k\underset{x\to a}{lim}f(x)+l\underset{x\to a}{lim}g(x)$$

which demonstrates the linearity of the limit operation. □

Theorem (Limit of a composite function)

Let $f(x)$ and $g(x)$ be functions such that $\underset{x\to a}{lim}f(x)=b$ and $\underset{x\to b}{lim}g(x)=\alpha$. Then

$$\underset{x\to a}{lim}(g\circ f)(x)=\alpha$$

where $g\circ f$ is the function composition of $g$ after $f$ (or $f$ then $g$), that is, $(g\circ f)(x)=g(f(x))$.

Proof. Let $\epsilon$ be an arbitrary positive real number. As usual, we can find ${\delta }_1>0$ such that $0<|x-b|<{\delta }_1$ implies $|g(x)-\alpha |<\epsilon$. Similarly, we can find $\delta >0$ such that $0<|x-a|<\delta$ implies $|f(x)-b|<{\delta }_1$. Then $0<|x-a|<\delta$ implies $|g(f(x))-\alpha |<\epsilon$. Hence

$$\underset{x\to a}{lim}g(f(x))=\alpha .$$

■

Example. If $x$ approaches $a$, then $x$ approaches $a$ (tautology!). Therefore

$$\underset{x\to a}{lim}x=a.$$

By the above theorems, we have, for example,

$$\begin{array}{rcl}\underset{x\to a}{lim}2x& =& 2a,\\ \underset{x\to a}{lim}{x}^2& =& a^2,\\ \underset{x\to a}{lim}(x^2+2x)& =& a^2+2a.\end{array}$$

In general, if $f(x)$ is a polynomial of $x$ (such as above), then

$$\underset{x\to a}{lim}f(x)=f(a).$$

□

Example. Consider

$$\underset{x\to 1}{lim}\frac{x^3-1}{x-1}.$$

Since we can factorize as $x^3-1=(x-1)(x^2+x+1)$, we have

$$\frac{x^3-1}{x-1}=\frac{(x-1)(x^2+x+1)}{x-1}=x^2+x+1$$

which is a polynomial of $x$. Therefore,

$$\underset{x\to 1}{lim}\frac{x^3-1}{x-1}=\underset{x\to 1}{lim}(x^2+x+1)=3.$$

□