Proof of Various Limit Properties — Question 8

Question 8 - Solution

We prove this by contradiction.

Assume, to the contrary, that $$L<0.$$

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

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

This inequality implies $$f(x)<L+\varepsilon = L+\frac{-L}{2} = \frac{L}{2}.$$

Because $L<0$, we have $$\frac{L}{2}<0.$$

Thus, for all $x$ with $0<|x-a|<\delta$, $$f(x)<0.$$

This contradicts the assumption that $f(x)\ge 0$ for all $x$ near $a$.

Therefore, our assumption that $L<0$ must be false. We conclude that $$L\ge 0.$$