Proof of Various Limit Properties — Question 4

Question 4 - Solution

Let $\varepsilon>0$. Since $g(x)\to M\ne0$, choose $\delta_1>0$ such that $|g(x)-M|<|M|/2$ when $0<|x-a|<\delta_1$. Then $|g(x)|>|M|/2$.

By the two assumed limits, choose $\delta_2,\delta_3>0$ to ensure respectively

$$|f(x)-L|<\frac{\varepsilon|M|}{4},\qquad |g(x)-M|<\frac{\varepsilon|M|^2}{4(1+|L|)}.$$

Take $\delta=\min(\delta_1,\delta_2,\delta_3)$. For $0<|x-a|<\delta$,

$$\begin{aligned} \left|\frac{f(x)}{g(x)}-\frac LM\right| &\leq\frac{|f(x)-L|}{|g(x)|}+\frac{|L|\,|g(x)-M|}{|M|\,|g(x)|}\\ &\leq\frac{2|f(x)-L|}{|M|}+\frac{2|L|\,|g(x)-M|}{|M|^2}\\ &<\frac\varepsilon2+\frac{\varepsilon|L|}{2(1+|L|)}<\varepsilon. \end{aligned}$$

(The first term makes the penultimate inequality strict even if $L=0$.) This proves $\displaystyle\lim_{x\to a}f(x)/g(x)=L/M$.