Notes - Analysis II HT23, L’Hôpital’s Rule

Created: March 22, 2023

Flashcards

Can you state the $\frac 0 0$ (or the $\frac{\infty}{\infty}$ form, as the conditions are equivalent) form of L’Hôpital’s Rule (for right sided limits)?

Suppose $f$ and $g$ are real-valued functions defined in some interval $(a, a+\delta)$ with $\delta > 0$. Furthermore, suppose that

  • $f, g$ are differentiable in $(a, a+\delta)$
  • $\lim _ {x \to a^+} f(x) = \lim _ {x \to a^+} g(x) = 0$.
  • $g’(x) \ne 0$ on $(a, a+\delta)$.
  • $\lim _ {x \to a^+} \frac{f’(x)}{g’(x)}$ exists Then
\[\lim_{x\to a^+} \frac{f(x)}{g(x)} = \lim_{x \to a^+} \frac{f'(x)}{g'(x)}\]

How can you prove that L’Hôpital’s Rule works for limits at infinty, i.e. in the case where $x \to \pm \infty$?

Substitute $y = \frac 1 x$.

When proving the $\frac \infty \infty$ case of L’Hôpital’s Rule for $f, g$ differentiable on $(a, a+\delta)$, what new variable do you introduce and why?

$a < x < c < a + \delta$, so that you can use the Cauchy’s Mean Value Theorem at a point that’s defined.

When proving the $\frac \infty \infty$ case of L’Hôpital’s Rule, you get to a stage where you have something like

\[|f(x) - f(c) - lg(x) + lg(c)| < \varepsilon |g(x) - g(c)|\]

How can you use the triangle inequality here to deduce something about $\frac{f(x)}{g(x)}$?
\[|f(x) - lg(x)| < \varepsilon |g(x) - g(c)| + |f(c) - lg(c)|\]

When proving the $\frac 0 0$ case of L’Hôpital’s Rule for $f, g$ differentiable on $(a, a+\delta)$, what two points do you apply the mean value theorem to?
\[x, a\]

If $a < \xi _ x < x$ and

\[\frac{f(x)}{g(x)} = \frac{f(x) - f(a)}{g(x) - g(a)} = \frac{f'(\xi_x)}{g'(\xi_x)}\]

what do you know about the limit of $\frac{f(x)}{g(x)}$ as $x \to a^+$ due to the limits of compositions of functions and the squeeze theorem?
\[\lim_{x\to a^+} \frac{f(x)}{g(x)} = \lim_{x\to a^+}\frac{f'(x)}{g'(x)}\]

What useful corollary is there to the $\frac 0 0$ form of L’Hôpital’s Rule, about $g(x)$?

$g(x)$ nonzero on $(a, a+\delta)$

Quickly prove the $\frac 0 0$ form of L’Hôpital’s Rule for right-sided limits, i.e.

Suppose $f$ and $g$ are real-valued functions defined in some interval $(a, a+\delta)$ with $\delta > 0$. Furthermore, suppose that

  • $f, g$ are differentiable in $(a, a+\delta)$
  • $\lim _ {x \to a^+} f(x) = \lim _ {x \to a^+} g(x) = 0$.
  • $g’(x) \ne 0$ on $(a, a+\delta)$.
  • $\lim _ {x \to a^+} \frac{f’(x)}{g’(x)}$ exists Then $g(x) \ne 0$ on $(a, a+\delta)$ and

We may assume/redefine $f(a) = g(a) = 0$ so that $f, g$ are continuous on $[a, a+\delta)$. Then for all $x \in (a, a+\delta)$, we have $g(x) = g(x) - g(a) \ne 0$ since otherwise Rolle’s theorem would imply $g’(x) \ne 0$, which we assume is not the case. Hence, by the generalised mean value theorem for all $x \in (a, a+\delta)$, we have that

\[\frac{f(x)}{g(x)} = \frac{f(x) - f(a)}{g(x)-g(a)} = \frac{f'(\xi_x)}{g'(\xi_x)}\]

for some $\xi _ x \in (a, a+\delta)$. Then, as $a < \xi _ x < x$, $\xi _ x \to x$ and so

\[\lim_{x\to a^+} \frac{f(x)}{g(x)} = \lim_{x \to a^+} \frac{f'(x)}{g'(x)}\]

When proving the $\frac \infty \infty$ form of L’Hôpital’s rule, can you justify that

\[g(x) - g(c) \ne 0\]

for all $a < x < c < a + \delta$?

Otherwise, by Rolle’s theorem there would exist a point in-between them with zero derivative, a contradiction to the assumptions.

Give the two “ingredients” that the rest of the $\frac \infty \infty$ form of L’Hôpital’s rule follows from: For $a < x < c < a + \delta$,

\[\left|\frac{f(x) - f(c)}{g(x) - g(c)}\right| = \left|\frac{f'(\xi_{x, c})}{g'(\xi_{x, c})}\right|\]

For any $\varepsilon$ and for $a < x < c < a + \delta’$ with some $\delta’ \in (0, \delta)$

\[\left|\frac{f(x) - f(c)}{g(x) - g(c)} - L \right| = \left|\frac{f'(\xi_{x, c})}{g'(\xi_{x, c})} - L \right| < \varepsilon\]

When proving L’Hopital’s rule, show that

\[\left|\frac{f(x) - f(c)}{g(x) - g(c)} - L \right| < \varepsilon \implies \left|\frac{f(x)}{g(x)} -L\right| < 2\varepsilon\]
\[\begin{aligned} &\left|\frac{f(x) - f(c)}{g(x) - g(c)} - L \right| < \varepsilon \\\\ \implies&|f(x) - f(c) - Lg(x) - Lg(c)| < \varepsilon|g(x) - g(c)| \\\\ \implies&|f(x) - Lg(x)| < \varepsilon |g(x) - g(c)| - |f(c) - Lg(c)| \\\\ \implies&\left|\frac{f(x)}{g(x)} -L\right| < \varepsilon \left|1 - \frac{g(c)}{g(x)}\right| - \left|\frac{f(c) - Lg(c)}{g(x)}\right| \\\\ \implies&\left|\frac{f(x)}{g(x)} -L\right| < 2\varepsilon \end{aligned}\]

with the second to last step a consequence of letting $x \to a$.

Proofs

Prove the $\frac 0 0$ form of L’Hôpital’s Rule for right-sided limits, i.e.

Suppose $f$ and $g$ are real-valued functions defined in some interval $(a, a+\delta)$ with $\delta > 0$. Furthermore, suppose that

  • $f, g$ are differentiable in $(a, a+\delta)$
  • $\lim _ {x \to a^+} f(x) = \lim _ {x \to a^+} g(x) = 0$.
  • $g’(x) \ne 0$ on $(a, a+\delta)$.
  • $\lim _ {x \to a^+} \frac{f’(x)}{g’(x)}$ exists Then $g(x) \ne 0$ on $(a, a+\delta)$ and
\[\lim_{x\to a^+} \frac{f(x)}{g(x)} = \lim_{x \to a^+} \frac{f'(x)}{g'(x)}\]

Todo.

Prove the $\frac \infty \infty$ form of L’Hôpital’s Rule for right-sided limits, i.e.

Suppose $f$ and $g$ are real-valued functions defined in some interval $(a, a+\delta)$ with $\delta > 0$. Furthermore, suppose that

  • $f, g$ are differentiable in $(a, a+\delta)$
  • $\lim _ {x \to a^+} f(x) = \lim _ {x \to a^+} g(x) = 0$.
  • $g’(x) \ne 0$ on $(a, a+\delta)$.
  • $\lim _ {x \to a^+} \frac{f’(x)}{g’(x)}$ exists Then $g(x) \ne 0$ on $(a, a+\delta)$ and
\[\lim_{x\to a^+} \frac{f(x)}{g(x)} = \lim_{x \to a^+} \frac{f'(x)}{g'(x)}\]

Todo.



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