Let's see if I can prove this by definition.
We are given that \(\displaystyle \lim_{x \to a}f(x) = A, \lim_{x \to a}g(x)=B\ and\ \lim_{x \to a}\dfrac{f(x)}{g(x)} = 1\)
\(\displaystyle So,\ given\ \epsilon >0,\ \exists\ \delta_1>0\ such\ that\ \mid f(x) - A \mid < \epsilon\ whenever\ 0< \mid x - a \mid \leq \delta_1 \)
\(\displaystyle and\ for\ the\ same\ \epsilon >0,\ \exists\ \delta_2>0\ such\ that\ \mid g(x) - B \mid < \epsilon\ whenever\ 0< \mid x - a \mid \leq \delta_2 \)
\(\displaystyle \lim_{x \to a}\dfrac{f(x)}{g(x)} = 1 \iff given\ any\ \epsilon>0, \exists\ \delta >0 such\ that\ \mid \dfrac {f(x)}{g(x)} - 1 \mid = \mid \dfrac {f(x)-g(x)}{g(x)} \mid< \epsilon\ whenever\ 0< \mid x - a \mid \leq \delta\)
Now if we take \(\displaystyle \delta\) smaller enough (at most \(\displaystyle \delta = min\){\(\displaystyle \delta_1, \delta_2\)}), g(x) will be with 1 of A. That is |g(x)-A|<1 (not the same 1 as the limit of f(x)/g(x))
Hence A-1<g(x)<A+1, so |g(x)| < max{|A-1|, |A+1|} = C
Now, \(\displaystyle \mid \dfrac {f(x)-g(x)}{C} \mid\ \leq\ \mid \dfrac {f(x)-g(x)}{g(x)} \mid < \epsilon\)
Putting this all together we have that given ϵ>0, ∃ δ>0 such that \(\displaystyle \mid f(x)-g(x) \mid\ < C\epsilon\ whenever\ 0< \mid x - a \mid \leq \delta\)
By definition of limit, this means that \(\displaystyle \lim_{x \to a}f(x)-g(x)=0,\ or\ that\ \lim_{x \to a}f(x)\ =\ \lim_{x \to a}g(x)\). Hence A=B.
I feel like a student right now by asking you nice helpers to please grade this proof, point out any any errors and how it can be improved?
Thanks!
