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New Zealand
Level 8 - NCEA Level 3

Limits and Continuity



Also important in calculus is the idea of continuity.  Basically, if we can trace the length of a function without taking our pencil off the paper, the function is continuous.

More formally

A function is continuous at $x=a$x=a if

(i) $f(a)$f(a) exists      

(ii)  $\lim_{x\rightarrow a}f\left(x\right)$limxaf(x) exists and equals $f(a)$f(a).



There are certain conditions which must exist for a function to be differentiable at $x=a$x=a

(i) The function is continuous at $x=a$x=a          

(ii) the gradient function is also continuous at $x=a$x=a

To say that the gradient function is continuous at $x=a$x=a is to say that the limit $\lim_{h\rightarrow0}\frac{f(a+h)-f(a)}{h}$limh0f(a+h)f(a)h exists. (Recall that the 'limit' needs to be the same whether the point $a$a is approached from above or from below.)

This limit may also be expressed as $\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$limxaf(x)f(a)xa.

We see that the notions of continuity and of differentiability are both dependent on the idea of limit.


It is usual to give one or more open intervals over which a function is continuous or differentiable. Open intervals are used because it is then possible to have a genuine limit at every point within the interval - where each point may be approached from both sides.

A notation $\lim_{x\rightarrow a^+}$limxa+ is used to indicate an approach to a point $a$a from above - a 'right-hand' limit. Similarly, $\lim_{x\rightarrow a^-}$limxa is used for a 'left-hand' limit.


Example 1

The function $g$g defined on the real numbers and given by $g(x)=x^2$g(x)=x2 is continuous on the open interval notated $(-\infty,\infty)$(,). It is also differentiable everywhere within this interval because the gradient function $g'(x)=2x$g(x)=2x is meaningful for all $x$x.


Example 2

The function $f$f defined on the set of real numbers without zero, given by $f(x)=\frac{1}{x}$f(x)=1x is discontinuous as $x=0$x=0 because $f(0)$f(0) does not exist. As a consequence, $f'(0)$f(0) does not exist. That is, the function is not differentiable at $x=0$x=0.

This function is, however, differentiable (and therefore continuous) on the intervals $(-\infty,0)$(,0) and $(0,\infty)$(0,).


Example 3

The rational function $h(x)=\frac{x^3-x^2}{x-1}$h(x)=x3x2x1 has a graph that looks perfectly smooth and continuous. It looks identical to the graph of the function given in Example 1, and yet the function must be undefined at $x=1$x=1 since the denominator would then be zero. 

If we were interested in the behaviour of this function between the values $x=-10$x=10 and $x=10$x=10, we might specify the domain as $\left[-10,1\right)\cup\left(1,10\right]$[10,1)(1,10].

The function is continuous and differentiable on the union of open intervals $\left(-10,1\right)\cup\left(1,10\right)$(10,1)(1,10)

In this case, the discontinuity at $x=1$x=1 can be 'removed' by defining the function value at $x=1$x=1 to be $1$1. If this is done, the function becomes continuous and differentiable over the whole interval $(-10,10)$(10,10).

Example 4

State the interval(s) of the domain over which the function is continuous.

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Example 5

Find $\lim_{x\to7^-}\left(\frac{\left|x-7\right|}{x-7}\right)$limx7(|x7|x7).

Example 6

Consider the function $f\left(x\right)=\frac{x}{\left(1-x\right)^3}$f(x)=x(1x)3.

  1. Find $\lim_{x\to1^+}f\left(x\right)$limx1+f(x).

  2. Find $\lim_{x\to1^-}f\left(x\right)$limx1f(x).





Identify discontinuities and limits of functions


Apply differentiation methods in solving problems

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