4. Convergence and continuity of a function

Limit of a function in a point

A very similar notion to the notion of limit of a sequence is the notion of limit of a function in a point. It describes the behaviour of values of the function when the arguments are nearer and nearer to a given number. There are two equivalent definitions of this notion: due to Heine and to Cauchy.

We will say that a function f has at a point x_0 limit g (denoted as \lim_{x\to x_0} f(x)=g), if for any sequence x_n which converges to x_0 and such that for any n, x_n\neq x_0, the limit of the sequence f(x_n) exists and equals g.

Simple example: function f(x)=x^2 has at 2 limit 4, because from the arithmetic of limits of sequences we know that if x_n converges to 2, then f(x_n)=x_n^2 converges to 2^2=4.

On the other hand, function

    \[g(x)=\begin{cases}-1, x<0\\ 1,x\geq 0\end{cases}\]

has no limit at 0, because for sequence x_n=\frac{1}{n+1}, f(x_n)\to 1, and for y_n=\frac{-1}{n+1}, f(y_n)\to -1.

Cauchy’s definition

The equivalent definition is the following. We will say that a function f has limit g at point x_0, if:

    \[\forall_{\varepsilon>0}\exists_{\delta}\forall_{x\in (x_0-\delta,x_0+\delta)\setminus\{x_0\}} |g-f(x)|<\varepsilon\]

which means that for arbitrarily positive small number \varepsilon there exists a small interval around x_0, such that values of the function in this small interval differ from the limit no more than \varepsilon.

Let us check, that lim_{x\to 0}x^2=0. Let \varepsilon>0. It suffices to set \delta=\sqrt{\varepsilon}. Then, if x\in(-\delta,\delta), then actually x^2<\delta^2=\varepsilon.

Infinite limits and limits at infinity

Using Heine’s definition we see that we can easily consider also limits of a function if x\to\infty lub x\to-\infty. Simply we can take all sequences x_n, which converge respectively to \infty or -\infty and look at the limits of sequences f(x_n). So for example, \lim_{x\to-\infty} \frac{x^3-x}{2x^3+1}=\lim_{x\to-\infty} \frac{1-\frac{1}{x^2}}{2+\frac{1}{x^3}}=\frac{1-0}{2-0}=\frac{1}{2}.

Also it can occur that the limit of a function at a given point is \infty or -\infty, if for all sequences x_n\to x_0 are such that the limit of f(x_n) is respectively \infty or -\infty. E.g. \lim_{x\to 0}\frac{1}{|x|}=\infty}.

Arithmetic of limits

The Heine’s version of definition immediately implies that the arithmetic of limits of functions works in the same way as the arithmetic of limits of sequences. E.g. \lim_{x\to -2}\frac{x^3+3x^2+6x+8}{x^2-4}=\lim_{x\to -2}\frac{(x+2)(x^2+x+4)}{(x+2)(x-2)}=\lim_{x\to -2}\frac{x^2+x+4}{x-2}=\frac{4-2+4}{-2-2}=\frac{6}{-4}=-3/2.

One-sided limits

Left limit (from below) of a function (denoted as \lim_{x\to\x_0^-} f(x)) equals g, if for any sequence x_{n} which converges to x_0, such that for any n, x_n<x_0 we get f(x_n)\to g. Similarly we define right limit (from above) (denoted as \lim_{x\to\x_0^+} f(x)) equals g, if for any sequence x_{n} which converges to x_0, such that for any n, x_n>x_0 we get f(x_n)\to g

E.g. \lim_{x\to 0^-} \frac{1}{x}=-\infty and \lim_{x\to 0^+} \frac{1}{x}=\infty.

A function has a limit at x_0 if and only if it has both limits in those points and they are equal.

Substitution theorem

We have the following theorem: \lim_{x\to x_0}f(g(x))=g if \lim_{y\to y_0}f(y)=g and \lim_{x\to x_0}g(x)=y_0, if for some neighbourhood of x_0, f(x)\neq y_0 for x\neq x_0.

Sounds a bit complicated but is very convenient. E.g. let us calculate the limit of function \frac{3^x-3}{9^x-3^x-6} for x=1. Let y(x)=3^x. If x\to 1, then y\to 3, and y(x)=3 if only x=1. Therefore:

    \[\lim_{x\to 1}\frac{3^x-3}{9^x-3^x-6}=\lim_{y\to 3}\frac{y-3}{y^2-y-6}=\lim_{y\to 3}\frac{y-3}{(y-3)(y+2)}=\lim_{y\to 3}\frac{1}{y+2}=\frac{1}{5}.\]


Asymptotes are lines which are lines to which the diagram of a function converges. Asymptotes can be vertical, horizontal or oblique.


If \lim_{x\to x_0^+} f(x)=\pm\infty, then line x=x_0 is a right vertical asymptote. Analogously, if \lim_{x\to x_0^-} f(x)=\pm\infty, this line is a left vertical asymptote. E.g. \lim_{x\to 0^+}\frac{1}{x}=\infty oraz \lim_{x\to 0^-}\frac{1}{x}=-\infty, so asymptote x=0 is a vertical asymptote of this function.

If \lim_{x\to\infty} f(x)=g or \lim_{x\to-\infty} f(x)=g, then line y=g is a respectively right or left horizontal asymptote of this function. Since \lim_{x\to\infty} 2^{-x}+2015=2015, y=2015 is a horizontal asymptote of 2^{-x}+2015=2015.

A line y=ax+b is an oblique asymptote (respectively left or right), if \lim_{x\to-\infty}f(x)-ax-b=0 or \lim_{x\to+\infty}f(x)-ax-b=0. If such a line is asymptote (assume it is a right asymptote), then a=\lim_{x\to\infty}\frac{f(x)}{x} oraz b=\lim_{x\to\infty}(f(x)-ax).

E.g.: let f(x)=\frac{2x^2-3}{3x-1}, then \lim_{x\to \infty}\frac{f(x)}{x}=\lim_{x\to \infty}\frac{2x^2-3}{3x^2-x}=\frac{2}{3} and \lim_{x\to\infty}\left(f(x)-\frac{2x}{3}\right)=\lim_{x\to\infty}\frac{\frac{2}{3}x-3}{3x-1}=\frac{2}{9}. Therefore, y=\frac{2}{3}x+\frac{2}{9} is an oblique asymptote of this function.


A function is continuous at a point x_0, if the limit \lim_{x\to x_0}f(x) exists and equals f(x_0). Obviously all simple arithmetic functions are continuous in every point of their domains. E.g. if f(x)=x^2, \lim_{x\to 2} f(x)= 4= f(2).

A function which is continuous at every point of its domain will be simply called continuous.

Function g(x)=|x| is continuous. Obviously it is continuous for all non-zero points. It is also continuous for x=0, because \lim_{x\to 0^+} |x|=\lim_{x\to 0^+} x = 0 and \lim_{x\to 0^-} |x|=\lim_{x\to 0^-} -x = 0, and therefore \lim_{x\to 0} |x| = 0=|0|.

Let h be a function defined in the following way:

    \[h(x)=\begin{cases} x^2\colon x\neq 2,\\ 2015\colon x=2\end{cases}.\]

This function is continuous in all points except 2, and is not continuous at 2, because \lim_{x\to 2} h(x)=4\neq 2015= h(2).

Let function r be defined as follows:

    \[r(x)=\begin{cases} -1\colon x<0,\\ 1\colon x\geq 0\end{cases}.\]

This function is continuous for x\neq 0, but is not continuous in 0, because \lim_{x\to 0^+} r(x)=1, but \lim_{x\to 0^-} r(x)=-1, and therefore it has no limit at point x=0.

Function w defined in the following way:

    \[w(x)=\begin{cases} 1\colon x\in\mathbb{Q},\\ 0\colon x\notin\mathbb{Q}\end{cases}.\]

is an example of function which has no limit in any point. Therefore is also not continuous at any point.

Darboux property

Continuous function on an interval [a,b] have the following (intuitively obvious) property: if f(a)<y<f(b), then there exists c, such that a<c<b, and f(c)=y. Therefore, for example, since for w(x)=-x^3+5x-1 is continuous and w(0)=-1 and w(1)=3, and therefore w(x) has at least one root in the interval (0,1).

Uniform continuity

If we are able to choose \delta for each \varepsilon, in a such way that in any interval of length \delta values of functions do not differ more than \varepsilon, universally regardless of place x, then we say that the function if uniformly continuous. More formally a function f is uniformly continuous, if:

    \[\forall_{\varepsilon>0}\exists_{\delta}\forall_{x_1,x_2} |x_1-x_2|\leq\delta\rightarrow |f(x_1)-f(x_2)|\leq\varepsilon.\]

Function f(x)=x is a very simple example. It is uniformly continuous, because for any \varepsilon set \delta=\varepsilon. Then for all x_1,x_2 such that |x_1-x_2|\leq\delta, then obviously |f(x_1)-f(x_2)|\leq\varepsilon.

On the other hand, function g(x)=\sin\frac{1}{x} on the interval (0,\infty) is continuous but is not uniformly continuous, because if \varepsilon=1, regardless how small \delta we choose, we can find 0<x_1<x_2<\delta, such that \sin\frac{1}{x_1}=-1, \sin\frac{1}{x_2}=-1, and therefore |f(x_1)-f(x_2)|=2>1.