Exercise 5.3.11

Question

\begin{enumerate}[label=(\alph*)] 
  \item Use the Generalized Mean Value Theorem to furnish a proof of the $0 / 0$ case of L'Hopital's Rule (Theorem 5.3.6).
  \item If we keep the first part of the hypothesis of Theorem 5.3.6 the same, but we assume that
    $$
    \lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}=\infty
    $$
    does it necessarily follow that
    $$
    \lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\infty ?
    $$
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item Let $f,g$ be continuous functions with $f(a) = g(a) = 0$ and $f'(x) e 0$, $g'(x) e 0$ around $a$, suppose $$ \lim_{x o a} \frac{f'(x)}{g'(x)} = L $$ We would like to show $\lim_{x o a} \frac{f(x)}{g(x)} = L$. Choose $\epsilon > 0$ then let $\delta > 0$ be such that $$ \left|\frac{f'(x)}{g'(x)} - L ight| < \epsilon $$ Let $x \in (a,a+\delta)$ and apply the generalized mean value theorem on $(a,x)$ to get a $c \in (a,x)$ with $$ \frac{f(x)-f(a)}{g(x)-g(a)} = \frac{f'(c)}{g'(c)} $$ Subtract $L$ from both sides and take absolute values, (and use $f(a) = g(a) = 0$) to get $$ \left|\frac{f(x)}{g(x)} - L ight| < \epsilon $$ We could do the same process starting from $x \in (a-\delta,a)$ as well, thus, for all $0 < |x| < \delta$ we have $$ \left|\frac{f(x)}{g(x)} - L ight| < \epsilon $$ Implying $\lim_{x o a} \frac{f(x)}{g(x)} = L$ as desired. An interesting thing to note is that \emph{the same $\delta$} works for both $f'(x)/g'(x)$ and for $f(x)/g(x)$. In other words, $f(x)/g(x)$ converges to $L$ at least as fast as $f'(x)/g'(x)$ does. \item Choose $M > 0$ and let $\delta > 0$ be such that $0<|x-a|<\delta$ implies $f'(x)/g'(x) > M$. Let $x \in (a,a+\delta)$ be arbitrary, then apply MVT on $(a,x)$ to get $c \in (a,x)$ with $$ \frac{f(x)}{g(x)} = \frac{f'(c)}{g'(c)} $$ Since $0<|c-a|<\delta$ we have $f'(c)/g'(c) > M$ and thus $$ \frac{f(x)}{g(x)} > M $$ for all $x \in (a,a+\delta)$, but again, we could just as easily apply this reasoning to $(a-\delta,a)$. So in general, all $x$ with $0 < |x-a| < \delta$ satisfy $$ \frac{f(x)}{g(x)} > M $$ Which is clearly the same as saying $\lim_{x o a} f(x)/g(x) = \infty$. \end{enumerate}

Exercise 5.3.11

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Exercise 5.3.11

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