Proof of Various Limit Properties — Question 7

Question 7 - Solution

We argue by contradiction.

Suppose, to the contrary, that $$L>M.$$

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 $$|f(x)-L|<\varepsilon.$$

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

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

This implies $$g(x)<M+\varepsilon=\frac{L+M}{2}.$$

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

Then for all $x$ satisfying $0<|x-a|<\delta$, we have $$f(x)>\frac{L+M}{2} \quad\text{and}\quad g(x)<\frac{L+M}{2}.$$

Thus, $$f(x)>g(x),$$ which contradicts the assumption that $f(x)\le g(x)$ near $a$.

Therefore, our assumption that $L>M$ must be false, and we conclude that $$L\le M.$$