Q 4. - Calculus

Question

If $\underset{x\to 0^{+}}{\lim }f\left(x\right)=-2$, $\underset{x\to 0^{-}}{\lim }f\left(x\right)=3$, and $f\left(0\right)=-2$, what can you say about $\underset{x\to 0}{\lim }f\left(x\right)$?

Step-by-Step Solution

Answer

$\underset{x\to 0}{\lim }f\left(x\right)$ does not exist.

Step 1. Given information.

It is given that $\underset{x\to 0^{+}}{\lim }f\left(x\right)=-2, \underset{x\to 0^{-}}{\lim }f\left(x\right)=3$ and $f\left(0\right)=-2$.

Step 2. Find lim x → 0 f ( x ) .

Here left-hand limit and the right-hand limit are not equal.

$\begin{array}{rcl}\underset{x\to 0^{+}}{\lim }f\left(x\right)& \ne & \underset{x\to 0^{-}}{\lim }f\left(x\right)\\ -2& \ne & 3\end{array}$

So, the limit of $f\left(x\right)$ as $x$ approaches to $0$ does not exist.

Step 3. Simplified answer.

Hence, $\underset{x\to 0}{\lim }f\left(x\right)$ does not exist.