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How to prove $\lim_{n\to\infty}\frac{f(x)}{g(x)} = \frac{\lim_{n\to\infty} f(x)}{\lim_{n\to\infty}g(x)}=\frac{L}{M}$ if g(x) is not equal to 0 using $\epsilon-\delta$ definition. I know the proof that uses the idea of $\frac{1}{g(x)}$ and the uses multiplication rule of limit, but I am wondering if there is a direct and more elegant proof.

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  • $\begingroup$ This proof is pretty direct, especially considering that division cannot be defined without a prior definition of multiplication (algebraically speaking). What do you have in mind exactly? $\endgroup$ – Gyu Eun Lee Apr 6 '14 at 19:16
  • $\begingroup$ You say "g(x) is not equal to 0" but you do not say that M, the limit of g(x), is not 0. That is separate and necessary. $\endgroup$ – user247327 Sep 28 '18 at 18:50
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Well, suppose both $f$ and $g$ converge ($g$ to something nonzero). Then we can make them as close to their limits as possible, so the difference $$\left|{f(x)\over g(x)}-{L\over M}\right|=\left|\frac{Mf(x)-Lg(x)}{Mg(x)}\right|=\left|\frac{M(f(x)-L)-L(g(x)-M)}{Mg(x)}\right|$$ can be made small by taking f and g sufficiently close to L and M, and by ensuring that g is far enough away from zero.

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