07Blimits

toc = Limits and continuity =

Notation
math \begin{matrix} \lim \\ x\to{a} \end{matrix} \;\; f(x) math

We read this as: "The limit of f(x) as x approaches a".

In other words, the limit is the value that f(x) approaches as x gets closer and closer to a.

** Example 1 **
math \text{Find } \begin{matrix} \lim \\ x\to3 \\ \end{matrix} \;\; (x + 3) math

__Solution:__

As x approaches 3 from the left (x → 3 – ) the value of the function approaches 6 (f(x) → 6). As x approaches 3 from the right (x → 3 + ) the value of the function approaches 6 (f(x) → 6).

The left hand limit is **equal** to the right hand limit. This can be written as: math \begin{matrix} \lim&(x+3)&=&\lim&(x+3)\\x\to3^-&\;&\;&x\to3^+&\; \end{matrix} math

The Limit at x = 3 exists if both the left hand limit and the right hand limit exist and are equal.

math \begin{matrix} \lim \\ x\to3 \\ \end{matrix} \;\; (x+3) = 6 math

f (x) = x + 3

f (3) = 6

Since the value of the function at the point where x = 3 is equal to the limit of the function as x approaches 3, the function is **continuous** at the point x = 3.

{This is the formal definition of continuity: see below}

**Example 2**
From the graph of the standard hyperbola we can observe the following limits:

math \begin{matrix} \lim\\ x\to +\infty \end{matrix} \;\; \dfrac{1}{x} = 0 math

math \begin{matrix} \lim \\ x\to-\infty \end{matrix} \;\; \dfrac{1}{x} =0 math

math \begin{matrix} \lim \\ x\to0^+ \end{matrix} \;\; \dfrac{1}{x} = +\infty math

math \begin{matrix} \lim \\ x\to0^- \end{matrix} \;\; \dfrac{1}{x} = -\infty math

**Example 3**
Determine the following limits from the graph:

math \textbf{(a) } \; \begin{matrix} \lim \\ x\to1^- \\ \end{matrix} \;\; \dfrac{1}{(x+2)(x-1)} math

math \textbf{(b) } \; \begin{matrix} \lim \\ x\to2^- \\ \end{matrix} \;\; \dfrac{1}{(x+2)(x-1)} math

math \textbf{(c) } \; \begin{matrix} \lim \\ x\to\infty \\ \end{matrix} \;\; \dfrac{1}{(x+2)(x-1)} math

__Solution:__

math \textbf{(a) } \; \begin{matrix} \lim \\ x\to1^- \\ \end{matrix} \;\; \dfrac{1}{(x+2)(x-1)} = -\infty math

math \textbf{(b) } \; \begin{matrix} \lim \\ x\to2^- \\ \end{matrix} \;\; \dfrac{1}{(x+2)(x-1)} = +\infty math

math \textbf{(c) } \; \begin{matrix} \lim \\ x\to\infty \\ \end{matrix} \;\; \dfrac{1}{(x+2)(x-1)} = 0 math

Other limits

**Example 4**

Consider the following hybrid function: math f(x) = \Bigg\{ \begin{matrix} x+1 & x \leqslant 1 \\ -x & x > 1 \\ \end{matrix} math

math \begin{matrix} \lim \\ x \to 1^- \\ \end{matrix} \quad f(x) = 2 math

math \begin{matrix} \lim \\ x \to 1^+ \\ \end{matrix} \quad f(x) = -1 math

math f(1)=2 math

At x = 1 the left hand limit is not equal to the right hand limit, therefore, (by the formal definition of continuity) the function is not continuous at x = 1.

Informal Definition of Continuity
The graph of a __continuous function__ can be drawn without lifting the pen off the paper.

The function is not continuous at a point where it is necessary to lift the pen to continue drawing.

**Example 5**
Sketch the graph of the hybrid function: math f(x)=\Bigg \{ \begin{matrix} 1,&x<0\\ x+2,&x\geqslant 0 \end{matrix} math

Use this graph to determine wheter //f(x)// is continuous at //x// = 0.

__Solution:__

math \begin{matrix} \lim \\ x\to0^- \end{matrix} \quad f(x) = 1 math

math \begin{matrix} \lim \\ x\to0^+ \\ \end{matrix} \quad f(x) = 2 math

Since the left hand limit is not equal to the right hand limit, the function is not continuous at x = 0. The break in the graph at x = 0 is a visual display of this discontinuity.

Determining limits algebraically
The limit, if it exists, can usually be found by substituting the value of the limit into the function.

If the function is not defined at that point, try factorising and cancelling terms before substituting.

**Example 6**
Evaluate the following limits:

math \textbf{(a) }\quad \begin{matrix} lim\\ x\to1 \end{matrix} \quad \dfrac{x+2}{x^2} math

math \textbf{(b) }\quad \begin{matrix} lim\\ x\to1 \end{matrix} \quad \dfrac{x-1}{x^2+2x-3} math

math \textbf{(c) }\quad \begin{matrix} lim\\ x\to2 \end{matrix} \quad \dfrac{x^3-8}{x-2} math

__Solution:__

math \\ \textbf{(a) }\quad\begin{matrix} lim\\ x\to1 \end{matrix} \quad\dfrac{x+2}{x^2} \\ \\ \\ .\qquad \qquad \quad =\dfrac{1+2}{1^2} \\ \\ .\qquad \qquad \quad =3 math

math \\ \textbf{(b) }\quad\begin{matrix} \lim\\ x\to1 \end{matrix} \quad\dfrac{x-1}{x^2+2x-3} \\ \\ \\ . \qquad \qquad \quad = \dfrac{1 - 1}{1 + 2 - 3} \\ \\ \\ . \qquad \qquad \quad = \dfrac{0}{0} math

{Substituting directly gives us "undefined" so try factorising and cancelling first!} math \\ . \qquad \qquad \quad = \begin{matrix} \lim \\ x\to1 \end{matrix} \quad \dfrac{x-1}{(x+3)(x-1)} \\ \\ \\ .\qquad \qquad \quad = \begin{matrix} lim \\ x\to1 \end{matrix} \quad \dfrac{1}{x+3} math

math \\ . \qquad \qquad \quad = \dfrac{1}{1+3} \\ \\ . \qquad \qquad \quad = \dfrac{1}{4} math

{The function is undefined at x = 1, therefore the graph is a hyperbola with a discontinuity at x = 1. }

math \\ \textbf{(c) }\quad \begin{matrix} \lim\\ x\to2 \end{matrix} \quad\dfrac{x^3-8}{x-2} \\ \\ \\ . \qquad \qquad \quad = \begin{matrix} \lim \\ x\to2 \end{matrix} \quad \dfrac{(x-2)(x^2+2x+4)}{x-2} math

math \\ . \qquad \qquad \quad = \begin{matrix} \lim \\ x\to2 \end{matrix} \quad x^2 + 2x + 4 \\ \\ . \qquad \qquad \quad = 2^2+2\times2+4 \\ \\ . \qquad \qquad \quad =12 math Go to top of page flat .