if $\lim_{x \to \infty} f(x)=L$ and $\lim_{x \to \infty}g(x)=M$ then $\lim_{x\to \infty}f(x)/g(x)=L/M$

Question

How can I prove that if $\lim_{x \to \infty} f(x)=L$ and $\lim_{x \to \infty}g(x)=M$ then $\lim_{x\to \infty}f(x)/g(x)=L/M$

What I did was to assume that

\begin{align*} \left|\frac{f(x)}{g(x)} - \frac{L}{M}\right| &= \left|\frac{f(x)}{g(x)} - \frac{L}{g(x)} + \frac{L}{g(x)} - \frac{L}{M}\right|\\ & = \left|\frac{f(x)}{g(x)} - \frac{L}{g(x)} + L \frac{M - g(x)}{Mg(x)}\right| \\ &\leq \left|\frac{1}{g(x)}\right| |f(x) - L| + \left|\frac{L}{M}\right| \left|\frac{1}{g(x)}\right| |M - g(x)|. \end{align*}

But I do not know how to continue the proof, since it is difficult for me to make a suitable choice of N such that x>N.

Answer 1

I will assume that $M\ne0$. Take $\varepsilon>0$. Then there is some $N_0\in\Bbb R$ such that$$x>N_0\implies|g(x)-M|<\frac{|M|}2\implies|g(x)|>\frac{|M|}2\iff\frac1{|g(x)|}<\frac2{|M|}.$$Now, take $N_1\in\Bbb R$ such that$$x>N_1\implies|f(x)-L|<\frac{\varepsilon|M|}4$$and take $N_2\in\Bbb R$ such that$$x>N_2\implies|g(x)-M|<\left|\frac ML\right|\cdot\frac{\varepsilon|M|}4.$$Then, if $N=\max\{N_0,N_1,N_2\}$, it follows from your inequality that$$x>N\implies\left|\frac{f(x)}{g(x)}-\frac LM\right|<\varepsilon.$$