Using the definition of the limit

Question

I need to prove that $\lim_{x\to 0}\frac{ax + b}{cx + d} = \frac{b}{d}$ using an , -style proof

Any ideas on how to do this?

Answer 1

Hint: consider difference $\left|\frac{ax+b}{cx+d}-\frac{b}{d}\right|$ and try to estimate, assuming, of course, natural conditions on fractions: $$\left|\frac{ax+b}{cx+d}-\frac{b}{d}\right| = \left|\frac{x(ad-bc)}{d(cx+d)}\right|= |x|\left|\frac{(ad-bc)}{d(cx+d)}\right|$$ For simplicity let's assume $c,d>0$, then, having $-\delta<x<\delta$ we obtain $-c\delta+d<cx+d<c\delta+d$. So, we can consider such $\delta>0$ for which $|cx+d|>M>0$ . Taking, for example $\delta >\frac{d}{2c}$ gives $|cx+d|>\frac{d}{2}$ So we will have $$\left|\frac{ax+b}{cx+d}-\frac{b}{d}\right| < \delta\frac{2|ad-bc|}{d^2}<\varepsilon$$. This is second condition on $\delta$ and joining with first finishes proof.