Proof of Various Limit Properties — Question 10

Question 10 - Solution

We prove this by contradiction.

Assume, to the contrary, that $$L\neq M.$$

Then the positive number $$\varepsilon=\frac{|L-M|}{2}$$ satisfies $\varepsilon>0$.

Since $\lim_{x\to a} f(x)=L$, there exists $\delta_1>0$ such that whenever $$0<|x-a|<\delta_1,$$ we have $$|f(x)-L|<\varepsilon.$$

Similarly, since $\lim_{x\to a} f(x)=M$, there exists $\delta_2>0$ such that whenever $$0<|x-a|<\delta_2,$$ we have $$|f(x)-M|<\varepsilon.$$

Let $$\delta=\min(\delta_1,\delta_2).$$

Then for all $x$ such that $0<|x-a|<\delta$, both inequalities hold.

Using the triangle inequality, $$|L-M| = |L-f(x)+f(x)-M| \le |f(x)-L|+|f(x)-M|.$$

Substitute the bounds: $$|L-M| < \varepsilon+\varepsilon =2\varepsilon =|L-M|.$$

This is a contradiction.

Therefore, our assumption that $L\neq M$ is false. We conclude that $$L=M.$$