Chapter 3.1, Problem 8QR

To determine

To calculate: The value of the limit $\underset{h\to 0^{+}}{\lim }f\left(1+h\right)$ .

Answer to Problem 8QR

The required value of the limit is 0.

Explanation of Solution

Given information:

The function: $f\left(x\right)=\{\begin{array}{cc}x+2,& x\le 1\\ {\left(x-1\right)}^2& x>1\end{array}$

Calculation:

It is seen that in the limit $\underset{h\to 0^{+}}{\lim }f\left(1+h\right)$ the term $h$ approaches zero from right.

This means the limit is same as the limit of the function where $x$ approaches 1 from right that is, $\underset{h\to 0^{+}}{\lim }f\left(1+h\right)=\underset{x\to 1^{+}}{\lim }f\left(x\right)$ .

For $x\to 1^{+}$ ; we can take the function $f\left(x\right)={\left(x-1\right)}^2$ .

So, $\begin{array}{l}\underset{x\to 1^{+}}{\lim }{\left(x-1\right)}^2={\left(1-1\right)}^2\\ =0\end{array}$

Thus, $\underset{h\to 0^{+}}{\lim }f\left(1+h\right)=0$

Hence, the required value of the limit is 0.