L'Hôpital's rule

-

We have been using the following formula without proof so far:

\[\lim_{x\to 0}\frac{\sin x}{x} = 1.\]

In this example, both \(\sin x\) and \(x\) converges to 0 as \(x \to 0\) so we have something like \(\frac{0}{0}\). In general, if \(\lim_{x\to a}f(x) = \lim_{x\to a}g(x) = 0\) or \(\lim_{x\to a}f(x) = \lim_{x\to a}g(x) = \pm\infty\), the limit of the form \(\lim_{x\to a}\frac{f(x)}{g(x)}\) is called an indeterminate form. L'Hôpital's rule provides a convenient way to calculate such limits (The proof will be given in another post).



See also: Proof of L'Hôpital's rule

Theorem (L'Hôpital's rule (1))

Let \(f(x)\) and \(g(x)\) be differentiable functions on the open interval \((a,b)\) that satisfy the following conditions.

  1. \[\lim_{x\to a+0}f(x) = \lim_{x\to a+0}g(x) = 0.\]
  2. For all \(x \in (a,b)\), \(g'(x) \neq 0\).
  3. The right limit \[\lim_{x\to a+0}\frac{f'(x)}{g'(x)}\] exists.
Then, the right limit \(\lim_{x\to a+0}\frac{f(x)}{g(x)}\) exists and
\[\lim_{x\to a+0}\frac{f(x)}{g(x)} = \lim_{x\to a+0}\frac{f'(x)}{g'(x)}.\]
Furthermore, condition 1 may be replaced with
  • (1') \[\lim_{x\to a+0}f(x) = \pm\infty \text{ and } \lim_{x\to a+0}g(x) = \pm\infty\]
to have the same conclusion.
Also the right limits \(\lim_{x \to a+0}\) may be replaced with the left limits \(\lim_{x\to a-0}\).

Corollary (L'Hôpital's rule (2))

Let \(f(x)\) and \(g(x)\) be differentiable functions on an open interval \(I\) that contains \(a\). Suppose \(f(x)\) and \(g(x)\) satisfy the following conditions.
  1. \[\lim_{x \to a}f(x) = \lim_{x\to a}g(x) = 0.\]
  2. For all \(x \in I\setminus\{a\}\), \(g'(x) \neq 0\).
  3. The limit \(\lim_{x\to a}\frac{f'(x)}{g'(x)}\) exists.
Then, the limit \(\lim_{x\to a}\frac{f(x)}{g(x)}\) exists and
\[\lim_{x\to a}\frac{f(x)}{g(x)} =\lim_{x\to a}\frac{f'(x)}{g'(x)}.\]
Furthermore, condition 1 may be replaced with
  • (1') \[\lim_{x\to a}f(x) = \pm\infty \text{ and } \lim_{x\to a}g(x) =\pm\infty\]
to have the same conclusion.

Example. Now let us prove
\[\lim_{x\to 0}\frac{\sin x}{x}  = 1\]
by applying L'Hôpital's rule. First, we need to check the conditions.
Both \(\sin x\) and \(x\) are defined on \(\mathbb{R}\) which is an open interval.
  1. We have \[\lim_{x\to 0} \sin x = 0\] and \[\lim_{x\to 0}x = 0.\]
  2. \((x)' = 1 \neq 0\) for all \(x\in \mathbb{R}\).
  3. We know \((\sin x)' = \cos x\) and \(\lim_{x\to 0}\cos x = 1\) so \[\lim_{x\to 0}\frac{(\sin x)'}{(x)'} = \lim_{x\to 0}\frac{\cos x}{1} = 1.\]
Therefore, all the conditions are satisfied and the limit \(\lim_{x \to 0}\frac{\sin x}{x}\) exists and
\[\lim_{x \to 0}\frac{\sin x}{x} = \lim_{x\to 0}\frac{(\sin x)'}{(x)'} = \lim_{x\to 0}\frac{\cos x}{1} = 1.\]
□

Example. Consider
\[\lim_{x\to 0}\frac{\log(\cos x)}{x^2}.\]
\(\lim_{x\to 0}\log(\cos x) = 0\) and \(\lim_{x\to 0}x^2 = 0\) and if \(x \neq 0\) then \((x^2)' = 2x \neq 0\). Furthermore,
\[ \begin{eqnarray} \lim_{x\to 0}\frac{\{\log(\cos x)\}'}{\{x^2\}'}&=& \lim_{x\to 0}\left(\frac{-\sin x}{\cos x}\cdot\frac{1}{2x}\right)\\ &=& \lim_{x\to 0}\left(\frac{\sin x}{x}\cdot\frac{-1}{2\cos x}\right)\\ &=& -\frac{1}{2}. \end{eqnarray} \]
Therefore,
\[\lim_{x\to 0}\frac{\log(\cos x)}{x^2} = -\frac{1}{2}.\]
□

Corollary (L'Hôpital's rule (3))

Let \(f(x)\) and \(g(x)\) be differentiable functions on the open interval \((b, \infty)\) that satisfy the following conditions.
  1. \[\lim_{x \to\infty}f(x) = \lim_{x\to\infty}g(x) = 0.\]
  2. For all \(x > b\), \(g'(x) \neq 0\).
  3. The limit \(\lim_{x\to \infty}\frac{f'(x)}{g'(x)}\) exists.
Then, the limit \(\lim_{x\to \infty}\frac{f(x)}{g(x)}\) also exists and
\[\lim_{x\to \infty}\frac{f(x)}{g(x)} = \lim_{x\to \infty}\frac{f'(x)}{g'(x)}.\]
Furthermore, condition 1 may be replaced with
  • (1') \[\lim_{x\to \infty}f(x) = \pm\infty \text{ and } \lim_{x\to \infty}g(x) =\pm\infty\]
to have the same conclusion.
Remark. The same corollary holds if \(f(x)\) and \(g(x)\) are continuous functions on \((-\infty, b)\) and the limits \(\lim_{x\to \infty}\) are replaced with \(\lim_{x \to -\infty}\). □
Example. Let us find \(\lim_{x\to\infty}x^{\frac{1}{x}}\).
Let \(f(x) = x^{\frac{1}{x}}\) so that \(\log f(x) = \frac{\log x}{x}\). Now consider \(\lim_{x\to\infty}\log f(x)\).
  1. \(\lim_{x\to\infty}\log x = \infty\) and \(\lim_{x\to\infty}x = \infty\).
  2. \((x)' = 1 \neq 0\).
  3. \[\lim_{x\to\infty}\frac{(\log x)'}{(x)'} = \lim_{x\to\infty}\frac{1}{x} = 0.\]
Thus,
\[\lim_{x\to\infty}\frac{\log x}{x} = 0.\]
Noting that \(f(x) = e^{\log f(x)}\) and that the exponential function is continuous, we find
\[\lim_{x\to\infty}x^{\frac{1}{x}} = \lim_{x\to\infty}e^{\frac{\log x}{x}} = e^0 = 1.\]
□

Comments

Post a Comment

Popular posts from this blog

Birth process

-
Image
  Defining the birth process  Consider a colony of bacteria that never dies. We study the following process known as  the birth process , also known as  the Yule process . The colony starts with \(n_0\) cells at time \(t = 0\). Assume that the probability that any individual cell divides in the time interval \((t, t + \delta t)\) is proportional to \(\delta t\) for small \(\delta t\). Further assume that each cell division is independent of others. Let \(\lambda\) be the  birth rate.  The probability of a cell division for a population of \(n\) cells during \(\delta t\) is \(\lambda n \delta t\). We assume that the probability that two or more births take place in the time interval \(\delta t\) is \(o(\delta t)\). That is, it can be ignored. Consequently, the probability that no cell divides during \(\delta t\) is \(1 - \lambda n \delta t - o(\delta t)\). Note that this process is an example of the Markov chain with states \({n_0}, {n_0 + 1}, {n_0 + 2}...
Read more

Branching processes: Mean and variance

-
Image
Generational growth Consider the following scenario (see the figure below): A single individual (cell, organism, etc.) produces \(j (= 0, 1, 2, \cdots)\) descendants with probability \(p_j\), independently of other individuals. The probability of this reproduction, \(\{p_j\}\), is known.  That individual produces no further descendants after the first (if any) reproduction. These descendants each produce further descendants at the next subsequent time with the same probabilities.  This process carries on, creating successive generations. Figure 1. An example of the branching process. Let \(X_n\) be the random variable representing the population size (number of individuals) of generation \(n\). In the above figure, we have \(X_0 = 1\), \(X_1=4\), \(X_2 = 7\), \(X_3=12\), \(X_4 = 9.\) We shall assume \(X_0 = 1\) as the initial condition. Ideally, our goal would be to find how the population size grows through generations, that is, to find the probability \(\Pr(X_n = k)\) for e...
Read more

Combining birth and death processes

-
Image
The birth-death process Combining birth and death processes with birth and death rates \(\lambda\) and \(\mu\), respectively, we expect to have the following differential-difference equations for the birth-death process : \[\begin{eqnarray}\frac{{d}p_0(t)}{{d}t} &=& \mu p_1(t),\\\frac{{d}p_n(t)}{{d}t} &=& \lambda(n-1)p_{n-1}(t) - (\lambda + \mu)np_n(t) + \mu(n+1)p_{n+1}(t),~~(n \geq 1).\end{eqnarray}\] You should derive the above equations based on the following assumptions: Given a population with \(n\) individuals, the probability that an individual is born in the population during a short period \(\delta t\) is \(\lambda n \delta t + o(\delta t)\). Given a population with \(n\) individuals, the probability that an individual dies in the population is \(\mu n \delta t + o(\delta t)\). The probability that multiple individuals are born or die during \(\delta t\) is negligible. (The probability of one birth and one death during \(\delta t\) is also negligible.) Consequ...
Read more