Chapter 8: Continuation of Limits

Limits

We briefly recall the notion of a limit.

$\lim_{x \rightarrow C} f(x) = L \Leftrightarrow$ $f(x)$ is close to $L$ when $x$ is close to $C$

Example

Calculate $\lim_{x \rightarrow 2} (-x^2+5x-2)$

Because the limit is a simple polynomial we may evaluate by substitution.

$\lim_{x \rightarrow 2} (-x^2+5x-2) = -(2)^2+5 \cdot 2 -2$

$=-4 +10 -2 = 4$

Example

Calculate $\lim_{z \rightarrow 4} \sqrt{z^2-10}$

$= \sqrt{4^2-10} = \sqrt{16-10}=\sqrt{6}$

We need to be careful with direct evaluation.

Example

Calculate $\lim_{x \rightarrow -3} \frac{x+3}{x^2+4x+3}$

$(-3)+3=0$ and $(-3)^2+4(-3)+3=0$

Thus we get "$\frac{0}{0}$" which is not a number

$\lim_{x \rightarrow -3} \frac{x+3}{x^2+4x+3}$ $=\lim_{x \rightarrow -3} \frac{x+3}{(x+4)(x+3)}$

$= \lim_{x \rightarrow -3} \frac{1}{x+4}$

$=\frac{1}{-3+4}$

$=-\frac{1}{2}$

One Sided Limits

Discuss

What is the behaviour of $f(x)=\sqrt{x}$ "close to" $x=0$?

Consider the graph:

1. Sloping up when $x \geq 0$
2. Not defined for $x<0$

Discuss

What is the behaviours of $f(x) = \frac{|x|}{x}$ "close tp" $x=0$?

Consider the graph:

1. $f(x)=1$ when $x \geq 0$
2. $f(x)=-1$ when $x<0$

We observe that there can be very distinct behaviour "on this left" and "on the right".

Definition

The right hand limit as $x$ approaches $x=c$ of $f(x)$ is

$\lim_{x \rightarrow C^+} f(x) = L$

The left hand limit is

$\lim_{x \rightarrow C^-} f(x) = L$

We obtain $\lim_{x \rightarrow 0^+} \frac{|x|}{x}=1$ and $\lim_{x \rightarrow 0^-} \frac{|x|}{x}=-1$

Example

Consider the following graph:

Which of the following exist (are defined)?

$\lim_{x \rightarrow -1^+} f(x)$ $\lim_{x \rightarrow -1^-} f(x)$

$\lim_{x \rightarrow 3^+} f(x)$ $\lim_{x \rightarrow 3^-} f(x)$

$\lim_{x \rightarrow 1^-} f(x)$ $\lim_{x \rightarrow 1^+} f(x)$

Definition:

The limit $\lim_{x \rightarrow C} f(x)$ EXISTS if $\lim_{x \rightarrow c^-} f(x) = \lim_{x \rightarrow c^+} f(x)$

Example

Pick a value $A$ so that :

$\lim_{x \rightarrow} f(x)$ exists when $f(x)=\left\{\begin{matrix} 3x+2 & x \geq 2\\ \ x^2+A& x<2 \end{matrix}\right.$

We need $\lim_{x \rightarrow 2^-} f(x) = \lim_{x \rightarrow 2^+} f(x)$

$\lim_{x \rightarrow 2^-} f(x) = \lim_{x \rightarrow 2^-} x^2 +A$ $\Rightarrow$ $x \rightarrow 2^-$ gives

$=4+A$

$\lim_{x \rightarrow 2^+}=\lim_{x \rightarrow 2^+} 3x+2$ $\Rightarrow$ $x \rightarrow 2^+$ gives $x \geq 2$

$=8$

Thus, $8=4+A$ and $A=4$