L'Hopital's Rule(s)

L'Hopital's Rule:

\frac{0}{0}

case

Let $f, g$ be continuous on an interval containing $a$ and suppose $f, g$ are differentiable on this interval with the possible exception of the point $a$ . If $lim_{x \to a} f (x) = lim_{x \to a} g (x) = 0$ and $g^{'} (x) \neq 0$ for all $x \neq a$ , then:

lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)} = L \in R \Rightarrow lim_{x \to a} \frac{f (x)}{g (x)} = L

Proof

Apply the Generalized Mean Value Theorem via the following method. For argument let $(b, c)$ be our interval.

Consider the case that $a = b$ . Let $ε > 0$ . Supposing $lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)} = L$ , choose $δ > 0$ such that $0 < | x - b | < δ$ implies that $| \frac{f^{'} (x)}{g^{'} (x)} - L | < \frac{ε}{2}$ . Let $x \in (b, b + δ)$ . Let $y \in (b, x)$ . By the Generalized Mean Value Theorem, then $\exists t \in (y, x)$ such that:

f^{'} (t) (g (x) - g (y)) = g^{'} (t) (f (x) - f (y))

and because $g (x) \neq g (y)$ and $g^{'} (t) \neq 0$ :

\frac{f^{'} (t)}{g^{'} (t)} = \frac{f (x) - f (y)}{g (x) - g (y)}

Now since $b < t < b + δ$ then our implication we started with gives:

L - \frac{ε}{2} < \frac{f (x) - f (y)}{g (x) - g (y)} < L + \frac{ε}{2}

for all $y \in (a, x)$ . Let $y \to b^{+}$ then by the Algebraic Limit Theorem for Functional Limits and Limits and Order (Order Limit Theorem) then taking a limit and using $lim_{x \to a} f (x) = 0 = lim_{x \to a} g (x)$ then:

L - ε < \frac{f (x)}{g (x)} < L + ε

Doing the cases for $a = c$ and $c \neq a, b$ are done very similarly.

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L'Hopital's Rule:

\frac{\infty}{\infty}

case

Suppose $f, g$ are differentiable on $(a, b)$ and that $g^{'} (x) \neq 0$ for all $x \in (a, b)$ . If $lim_{x \to a} g (x) = \infty$ (defined here) (or $- \infty$ ), then:

lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)} = L \Rightarrow lim_{x \to a} \frac{f (x)}{g (x)} = L

Notice that we only need the denominator to go to

\infty

in this case. This is different than the

\frac{0}{0}

case.

Proof

Let $ε > 0$ . Because $lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)} = L$ then $\exists δ_{1} > 0$ such that:

| \frac{f^{'} (x)}{g^{'} (x)} - L | < \frac{ε}{2}

for all $a < x < a + δ_{1}$ . For convenience of notation, let $t = a + δ_{1}$ and note that $t$ is fixed for the remainder of the argument.

Our functions are not defined at $a$ , but for any $x \in (a, t)$ we can apply the Generalized Mean Value Theorem on the interval $[x, t]$ to get:

\frac{f (x) - f (t)}{g (x) - g (t)} = \frac{f^{'} (c)}{g^{'} (c)}

for some $c \in (x, t)$ . Our choice of $t$ then implies that:

L - \frac{ε}{2} < \frac{f (x) - f (t)}{g (x) - g (t)} < L + \frac{ε}{2}

for all $x \in (a, t)$ .

In an effort to isolate the fraction $\frac{f (x)}{g (x)}$ , the strategy is to multiply the inequality by $\frac{g (x) - g (t)}{g (x)}$ . We need to be sure, however, that this quantity is positive, which amounts to insisting that $1 \geq \frac{g (t)}{g (x)}$ . Now, because $t$ is fixed and $lim_{x \to a} g (x) = \infty$ then we can choose $δ_{2} > 0$ so that $g (x) \geq g (t)$ for all $a < x < a + δ_{2}$ (because $t$ is fixed, so eventually we pass $g (t)$ as $g (x) \to \infty$ ). Carrying out the desired multiplication results in:

(L - \frac{ε}{2}) (1 - \frac{g (t)}{g (x)}) < \frac{f (x) - f (t)}{g (x)} < (L + \frac{ε}{2}) (1 - \frac{g (t)}{g (x)})

which after some manipulation becomes:

L - \frac{ε}{2} + \frac{- L g (t) + \frac{ε}{2} g (t) + f (t)}{g (x)} < \frac{f (x)}{g (x)} < L + \frac{ε}{2} + \frac{- L g (t) - \frac{ε}{2} g (t) + f (t)}{g (x)}

Notice the $g (x)$ 's in the denominator. Again, because $t$ is fixed then $lim_{x \to a} g (x) = \infty$ . Thus, we can choose a $δ_{3}$ such that $x \in (a, a + δ_{3})$ implies that $g (x)$ is large enough to ensure that the terms:

\frac{- L g (t) + \frac{ε}{2} g (t) + f (t)}{g (x)}, \frac{- L g (t) - \frac{ε}{2} g (t) + f (t)}{g (x)}

are less than $\frac{ε}{2}$ in absolute value (because the numerators are fixed, and $g (x) \to \infty$ ), so using that in our original inequality and choosing $δ = min {δ_{1}, δ_{2}, δ_{3}}$ guarantees that:

L - \frac{ε}{2} + \frac{ε}{2} < \frac{f (x)}{g (x)} < L + \frac{ε}{2} + \frac{ε}{2}

Thus:

| \frac{f (x)}{g (x)} - L | < ε

for all $a < x < a + δ$ .

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