Exercise 5.3.9

Question

Assume $f$ and $g$ are as described in Theorem 5.3.6, but now add the assumption that $f$ and $g$ are differentiable at $a$, and $f^{\prime}$ and $g^{\prime}$ are continuous at $a$ with $g^{\prime}(a) 
eq 0$. Find a short proof for the $0 / 0$ case of L'Hopital's Rule under this stronger hypothesis.

Ulisse Mini · Accepted Answer

Let $(x_n)$ be a sequence approaching $a$ and apply MVT on $[x_n,a]$ to find $c_n,d_n \in (x_n,a)$ with
  $$
  f'(c_n) = \frac{f(x_n)}{x_n - a} \quad	ext{and}\quad g'(d_n) = \frac{g(x_n)}{x_n - a}
  $$
  Meaning
  $$
  \lim \frac{f(x_n)}{g(x_n)} = \lim \frac{f'(c_n)/(x_n-a)}{g'(d_n)/(x_n-a)} = \lim \frac{f'(c_n)}{g'(d_n)}
  $$
  The continuity of $f'$ and $g'$ combined with $g'(a) 
e 0$ implies the limit exists
  $$
  \lim \frac{f'(c_n)}{g'(d_n)} = \frac{f'(a)}{g'(a)} = L
  $$
  Since we showed $\lim f(x_n)/g(x_n) = L$ \emph{for all} sequences $(x_n)$ the Sequential Criterion for Functional Limits (Theorem 4.2.3) implies
  $$
  \lim_{x 	o a} \frac{f(x)}{g(x)} = L
  $$
  Which completes the proof.

Exercise 5.3.9

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Exercise 5.3.9

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