Mathematics Asked by Ashish on January 6, 2021
Suppose that, $limlimits_{xto x_0}f(x)$ and $limlimits_{xto x_0}g(x)$ exist and $limlimits_{xto x_0}g(x)neq0$. Then, prove that $$limlimits_{xto x_0}frac{f(x)}{g(x)}=frac{limlimits_{xto x_0}f(x)}{limlimits_{xto x_0}g(x)}$$
Let $K$ and $L$ be the limits of $f$ and $g$ respectively. Let $epsilon > 0 $ and choose $delta > 0 $ such that for $|x-x_0| < delta$ we have $|f(x) - K| < epsilon$ and $|g(x) - L| < epsilon.$ We can also conveniently suppose that in this range $|g(x)| geq |L|/2 .$ Then
begin{align} |f(x)/g(x) - K/L| &= bigg|frac{f(x)L - Kg(x)}{Lg(x)}bigg| \ &= bigg|frac{f(x)L - KL + KL - Kg(x)}{Lg(x)}bigg| \ &= bigg|frac{L(f(x) - K) + K(L - g(x))}{Lg(x)}bigg| \ &< epsilon(L+K)/L|g(x)|\ &leq epsilon2(L+K)/L^2. end{align} Of course if you want all of this to be neatly less than $epsilon$ then you could change your values of $delta$ accordingly.
Answered by William O'Regan on January 6, 2021
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