Lecture 8: One Sided Limits

To have a limit $L$ as $x$ approaches $c$, the function must be defined on both sides of c and the $f(x)$ values must approach $L$

$\rightarrow$ Two sides limits

Functions can have one sided limit if approach from one side

$\lim_{x \rightarrow c^+} f(x) = L$ $\Rightarrow$ A right hand limit

$\lim_{x \rightarrow c^-} f(x) = L$ $\Rightarrow$ A left hand limit

Rule

The limit exits if a function $f(x)$ has a limit as $x$ approaches $c$. If it has left handed and right handed limits and both are equal

$\lim_{x \rightarrow c} f(x) = L$

If $\lim_{x \rightarrow c^-} f(x) = L=\lim_{x \rightarrow c^+} f(x)$

Rule

Also remember this side rule to help solve problem

$\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$, where $\theta$ can be replaced by $x$

Example 1

Does the limit exist for $\lim_{x \rightarrow 3} \frac{|x-3|}{x-3}$?

$RHL \rightarrow \lim_{x \rightarrow 3^+} \frac{|x-3|}{x-3} = \lim_{x \rightarrow 3^+} \frac{x-3}{x-3} = \frac {x-3}{x-3}=1$

$LHL \rightarrow \lim_{x \rightarrow 3^-} \frac{|x-3|}{x-3} = \lim_{x \rightarrow 3^-} \frac{-x-3}{x-3} = \frac {-(x-3)}{x-3}=-1$

$\Rightarrow 1 \neq 1 \therefore$ Limit DNE

EXAMPLE 2

$\lim_{x \rightarrow 0} \frac{\sin 3x}{4x} = \lim_{x \rightarrow 0} \frac{\sin 3x}{4/3 \cdot 3x}= \frac{3}{4} \lim_{x \rightarrow 0} \frac{\sin 3x}{3x} = \frac{3}{4} \cdot 1 = \frac{3}{4}$