Chapter A3, Problem 18E

To determine

To prove: The Difference rule.

Explanation of Solution

Proof:

Prove the difference rule for limits.

  $|f\left(x\right)-g\left(x\right)-\left(L-M\right)|=|f\left(x\right)-L+M-g\left(x\right)|\le |f\left(x\right)-L|+|g\left(x\right)-M|$

Since $\underset{x\to c}{\lim }f\left(x\right)=L$

  $\underset{x\to c}{\lim }g\left(x\right)=M$

For any $\epsilon$ , there exists ${\delta }_1$ , ${\delta }_2$ such that

  $\begin{array}{l}0<|x-c|<{\delta }_1\Rightarrow |f\left(x\right)-L|<\frac{\epsilon }{2}and\\ 0<|x-c|<{\delta }_2\Rightarrow |g\left(x\right)-M|<\frac{\epsilon }{2}.\end{array}$

Choose $\delta =\min \left\{{\delta }_1,{\delta }_2\right\}$ .

Then $0<|x-c|<\delta$

  $\begin{array}{l}\Rightarrow |f\left(x\right)-L|+|g\left(x\right)-M|<\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon \\ \Rightarrow |f\left(x\right)-g\left(x\right)-\left(L-M\right)|<\epsilon \end{array}$

Conclusion:

Therefore, the given Difference rule for limits is proved.