Chapter A3, Problem 13E

To determine

To prove: The given limit.

Answer to Problem 13E

Here, the given $\underset{x\to 1}{\lim }f\left(x\right)=1$ is proved.

Explanation of Solution

Given: $f\left(x\right)=\{\begin{array}{l}{x}^2,x\ne 1\\ 2,x=1\end{array}$

Calculation:

Step 1:

  $\begin{array}{l}Forx\ne 1,|x^2-1|<\epsilon \\ \Rightarrow -\epsilon <x^2-1<\epsilon \\ \Rightarrow 1-\epsilon <x^2<1+\epsilon \\ \Rightarrow \sqrt{1-\epsilon }<x<\sqrt{1+\epsilon }\end{array}$

Since $x$ is near $1$ , use positive values only.

Step 2:

  $\begin{array}{l}|x-1|<\delta \\ \Rightarrow -\delta <x-1<\delta \\ \Rightarrow 1-\delta <x<1+\delta \\ Then1-\delta =\sqrt{1-\epsilon }\\ \Rightarrow \delta =1-\sqrt{1-\epsilon }\\ or1+\delta =\sqrt{1+\epsilon }\\ \Rightarrow \delta =\sqrt{1-\epsilon }-1\end{array}$

Choose $\delta =\min \left\{1-\sqrt{1-\epsilon },\sqrt{1+\epsilon }-1\right\}$ .

Conclusion:

Therefore, the given $\underset{x\to 1}{\lim }f\left(x\right)=1$ is proved.