More on the limit of functions

-

 There are a few variations of limits.



Definition (Divergence of a function)

If the function \(f(x)\) does not converge to any value as \(x \to a\), we say that \(f(x)\) diverges as \(x \to a\). In particular, if the value of \(f(x)\) increases arbitrarily as \(x \to a\), we say that \(f(x)\) diverges to positive infinity and write

\[\lim_{x\to a}f(x) = \infty.\]

If the value of \(f(x)\) is negative and its absolute value increases arbitrarily, we say that \(f(x)\) diverges to negative infinity and write

\[\lim_{x\to a}f(x) = -\infty.\]

Example. Consider the function

\[f(x) = \frac{1}{(x - 1)^2}\]

defined on \(\mathbb{R}\setminus\{1\}\). We have

\[\lim_{x \to 1}f(x) = \infty.\]

You should verify this by drawing a graph. □

There are a few variants of the notion of limit.

Definition (Left and right limits)

  1. We define the left limit of the function \(f(x)\) at \(x=a\) \[\lim_{x \to a -0}f(x) = \alpha\] if the following is satisfied
    • For any \(\varepsilon > 0\), there exists \(\delta > 0\) such that for any \(x\in\text{dom}(f)\), if \(0 < a - x < \delta\) then \(|f(x) - \alpha| < \varepsilon\)
    This mean \(x\) approaches \(a\) from the left (so ``\(x < a\)'' is kept).
  2. We define the right limit of the function \(f(x)\) at \(x=a\) \[\lim_{x \to a +0}f(x) = \alpha\] if the following is satisfied
    • For any \(\varepsilon > 0\), there exists \(\delta > 0\) such that for any \(x\in\text{dom}(f)\), if \(0 < x - a < \delta\) then \(|f(x) - \alpha| < \varepsilon\).
    This mean \(x\) approaches \(a\) from the right (so ``\(a < x\)'' is kept).
Remark. We write \(x \to +0\) instead of \(x \to 0+0\). Similarly \(x\to -0\) rather than \(x \to 0-0\). □

Example. Consider the function 
\[f(x) = \frac{x(x + 1)}{|x|}\]
defined on \(\mathbb{R}\setminus\{0\}\). 
If \(x > 0\), we have
\[f(x) = \frac{x(x + 1)}{x} = x + 1\]
so that
\[\lim_{x \to +0}f(x) = 1.\]
If \(x < 0\), we have
\[f(x) = \frac{x(x + 1)}{-x} = -x - 1\]
so that
\[\lim_{x \to -0}f(x) = -1.\]
□

Example. For the function \(f(x) = \frac{1}{x}\) defined on \(\mathbb{R}\setminus\{0\}\), we have
\[ \begin{eqnarray*} \lim_{x\to +0}f(x) &=& \infty,\\ \lim_{x\to -0}f(x) &=& -\infty. \end{eqnarray*} \]
□

Definition (Limits at \(\pm \infty\))

  1. We define the limit at \(x \to \infty\) \[\lim_{x \to \infty}f(x) = \alpha\] if the following is satisfied
    • For any \(\varepsilon > 0\), there exists \(\delta > 0\) such that for any \(x\in\text{dom}(f)\), if \(x > \delta\) then \(|f(x) - \alpha| < \varepsilon\).
  2. We define the limit at \(x \to -\infty\) \[\lim_{x \to -\infty}f(x) = \alpha\] if the following is satisfied
    • For any \(\varepsilon > 0\), there exists \(\delta > 0\) such that for any \(x\in\text{dom}(f)\), if \(x < -\delta\) then \(|f(x) - \alpha| < \varepsilon\).

Theorem

Let \(f(x)\) be a function defined on the open interval \(I = (a, b)\) \((a < b)\) such that \(\lim_{x\to a+0}f(x) = \alpha\) \((\alpha \in\mathbb{R})\).
  1. For a given \(c\in\mathbb{R}\), if \(f(x) \geq c\) for all \(x\in I\), then \(\alpha \geq c\).
  2. For a given \(d\in\mathbb{R}\), if \(f(x) \leq d\) for all \(x\in I\), then \(\alpha \leq d\).
Proof. We prove only part 1 and prove it by contradiction. Part 2 is similar.
Suppose \(\alpha < c\) and let \(\varepsilon = c - \alpha > 0\). Since \(\lim_{x\to a+0}f(x) = \alpha\), we can find \(\delta > 0\) such that \(0 < x - a < \delta\) implies \(|f(x) - \alpha| < \varepsilon\). But then \(f(x) < \alpha + \varepsilon = c\), which is a contradiction. ■

Theorem

Let \(f(x)\) be a function defined on an interval that includes \(x= a\).
  1. If \(\lim_{x\to a}f(x) = \alpha\) exists, then the corresponding left and right limits also exist and \(\lim_{x\to a+0}f(x) = \lim_{x\to a-0}f(x) = \alpha\).
  2. If both \(\lim_{x\to a+0}f(x)\) and \(\lim_{x\to a-0}f(x)\) exist and \(\lim_{x\to a+0}f(x) = \lim_{x\to a-0}f(x) = \alpha\), then \(\lim_{x\to a}f(x) = \alpha\).
Proof. Exercise. ■

Example. Let us prove the following:
\[e = \lim_{x\to 0}(1 + x)^{\frac{1}{x}} = \lim_{x\to\infty}\left(1 + \frac{1}{x}\right)^x  = \lim_{x\to -\infty}\left(1 + \frac{1}{x}\right)^x.\]
Before proving this, recall that we have
\[\lim_{n\to \infty}\left(1 + \frac{1}{n}\right)^n = e\]
where \(n = 1, 2, 3, \cdots\) (i.e., natural numbers).

In order to show \(e = \lim_{x\to 0}(1 + x)^{\frac{1}{x}}\), we need to show \(e = \lim_{x\to +0}(1 + x)^{\frac{1}{x}}\) and \(e = \lim_{x\to -0}(1 + x)^{\frac{1}{x}}\). By changing variables \(x = \frac{1}{y}\), \(\lim_{x\to +0}(1 + x)^{\frac{1}{x}} = \lim_{y\to\infty}\left(1 + \frac{1}{y}\right)^y\) and \(\lim_{x\to -0}(1 + x)^{\frac{1}{x}} = \lim_{y\to -\infty}\left(1 + \frac{1}{y}\right)^y\). Furthermore,  by changing variables \(y = -t\), assuming \(t  > 1\), we have
\[ \begin{eqnarray*} \lim_{y\to -\infty}\left(1 + \frac{1}{y}\right)^y &=& \lim_{t\to \infty}\left(1 - \frac{1}{t}\right)^{-t}\\ &=& \lim_{t\to \infty}\left(\frac{t}{t-1}\right)^{t}\\ &=& \lim_{t\to \infty}\left(\frac{t}{t-1}\right)^{t-1}\left(\frac{t}{t-1}\right). \end{eqnarray*} \]
Since \(\lim_{t\to\infty}\left(1 + \frac{1}{t-1}\right) = 1\), we can see that we only need to show \(e = \lim_{x\to\infty}\left(1 + \frac{1}{x}\right)^{x}\).

For any \(x \geq 1\), we can find \(n\in\mathbb{N}\) such that
\[n\leq x \leq n+1.\]
Then we have
\[1 < 1 + \frac{1}{n+1} \leq 1 + \frac{1}{x} \leq 1 + \frac{1}{n}.\]
And hence,
\[\left(1 + \frac{1}{n+1}\right)^{n} \leq \left(1 + \frac{1}{x}\right)^{x} \leq \left(1 + \frac{1}{n}\right)^{n+1}.\]
Now,
\[\lim_{n\to\infty}\left(1 + \frac{1}{n+1}\right)^{n} = \lim_{n\to\infty}\frac{\left(1 + \frac{1}{n+1}\right)^{n+1}}{\left(1 + \frac{1}{n+1}\right)} = e,\]
and
\[\lim_{n\to\infty}\left(1 + \frac{1}{n}\right)^{n+1} = \lim_{n\to\infty}\left(1 + \frac{1}{n}\right)^{n}\left(1 + \frac{1}{n}\right) = e.\]
Therefore, by the Squeeze Theorem,
\[\lim_{x\to\infty}\left(1 + \frac{1}{x}\right)^{x} = e.\]
□

In a previous post, we have seen Cauchy's theorem for convergence of sequences (i.e., a Cauchy sequence converges and vice versa). 
A similar result hold for the convergence of functions.

Theorem (Cauchy criterion for the right limit of a function)

Let \(f(x)\) be a function on \(I = (a,b]\). The right limit \(\lim_{x\to a+0}f(x)\) exists if and only if the following condition is satisfied:
  • For any \(\varepsilon > 0\), there exists a \(\delta > 0\) such that for all \(x, y\in I\), if \(a < x < a + \delta\) and \(a < y < a + \delta\) then \(|f(x) - f(y)| < \varepsilon\).     (\(\star\))
In a logical form:
\[\forall \varepsilon > 0, \exists \delta > 0, \forall x,y\in I ~ (x,y \in (a,a+\delta) \implies |f(x) - f(y)|< \varepsilon).\]
Remark. Similarly, the left limit \(\lim_{x\to b-0}f(x)\) exists if and only if
  • For any \(\varepsilon > 0\), there exists a \(\delta > 0\) such that for all \(x, y\in I\), if \(b - \delta < x < b\) and \(b -\delta < y < b\) then \(|f(x) - f(y)| < \varepsilon\).
The limit \(\lim_{x\to\infty}f(x)\) exists if and only if
  • For any \(\varepsilon > 0\), there exists a \(\delta > 0\) such that for all \(x, y\in I\), if \(x > \delta\) and \(y > \delta\) then \(|f(x) - f(y)| < \varepsilon\).
The limit \(\lim_{x\to-\infty}f(x)\) exists if and only if
  • For any \(\varepsilon > 0\), there exists a \(\delta > 0\) such that for all \(x, y\in I\), if \(x < -\delta\) and \(y < -\delta\) then \(|f(x) - f(y)| < \varepsilon\).
□
Proof. (\(\Rightarrow\)) Suppose \(\lim_{x\to a+0}f(x)\) exists. Then there exists some real number \(\alpha\) such that \(\lim_{x\to a+0}f(x) = \alpha\). For any \(\varepsilon > 0\), there exists \(\delta > 0\) such that, for all \(x\in I\), if \(a < x < a + \delta\) then \(|f(x) - \alpha| < \frac{\varepsilon}{2}\). Let \(x\) and \(y\) be arbitrary real numbers such that \(a < x < a + \delta\) and \(a < y < a + \delta\). Then
\[ \begin{eqnarray*} |f(x) - f(y)| & = & |f(x) - \alpha + \alpha - f(y)|\\ &\leq &|f(x) - \alpha| + |f(y) - \alpha|\\ &<& \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{eqnarray*} \]
Therefore the condition (\(\star\)) is satisfied.

(\(\Leftarrow\)) Suppose (\(\star\)) is satisfied. Let us define a sequence \(\{a_n\}\) by \(a_n = a + \frac{b-a}{n}\), \(n\in\mathbb{N}\). For any \(\varepsilon > 0\),  choose a \(\delta > 0\) satisfying (\(\star\)). By Archimedes' principle, we can find an \(N\in \mathbb{N}\) such that \(N\delta > b - a\). Then for any \(n \geq N\), \(a_n - a  = \frac{b-a}{n} < \frac{b-a}{N} < \delta\) so that \(a < a_n < a + \delta\). Therefore by (\(\star\)), for all \(n, m > N\), \(|f(a_n) - f(a_m)| < \varepsilon\) so that \(\{f(a_n)\}\) is a Cauchy sequence, and hence this sequence converges: \(\lim_{n\to\infty}f(a_n) = \alpha\) for some \(\alpha \in \mathbb{R}\).

Now, let us show that \(\lim_{x\to a+0}f(x) = \alpha\). By (\(\star\)), for any \(\varepsilon > 0\), we can find \(\delta > 0\) such that for all \(x,y \in (a, a+\delta)\), \(|f(x) - f(y)|< \frac{\varepsilon}{2}\). Also, since \(\lim_{n\to\infty}f(a_n) = \alpha\), we can find an \(N\in\mathbb{N}\) such that for all \(n \geq N\), \(|f(a_n) - \alpha| < \frac{\varepsilon}{2}\). If \(a < x < a + \delta\), choose \(n \geq N\) such that \(a < a_n < a+\delta\). Then
\[ \begin{eqnarray*} |f(x) - \alpha| &=& |f(x) - f(a_n) + f(a_n) - \alpha|\\ &\leq&|f(x) - f(a_n)| + |f(a_n) - \alpha|\\ &<& \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon. \end{eqnarray*} \]
Therefore,
\[\lim_{x\to a+0}f(x) = \alpha.\]
■


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

Computing integrals (3): Integration by parts

-
Image
 Sometimes, we may simplify integration by using the product rule of differentiation. This technique is called integration by parts. Theorem (Integration by parts) Let \(f(x)\) and \(g(x)\) be differentiable functions on an open interval \(I\). Then,  \(\int f(x)g'(x)dx = f(x)g(x) - \int f'(x)g(x)dx\); For any \(a, b \in I\), \[\int_a^bf(x)g'(x)dx = \left[f(x)g(x)\right]_a^b - \int_a^bf'(x)g(x)dx.\] Proof . By the product rule, \[[f(x)g(x)]' = f'(x)g(x) + f(x)g'(x)\] so \[f(x)g'(x) = [f(x)g(x)]' - f'(x)g(x).\] By integrating both sides, we have the desired results. ■ Example . Let us find \(\int x\cosh x dx\). \[ \begin{eqnarray*} \int x\cosh x dx &=& \int x(\sinh x)'dx \\ &=& x \sinh x - \int 1 \cdot \sinh x dx\\ &=& x \sinh x - \cosh x + C. \end{eqnarray*} \] Example (eg:recur) . Let us study how we can compute \[I_n = \int \frac{dx}{(x^2 + 1)^n}\] for \(n\in \mathbb{N}\). Note \[I_{n} = \int \fr...
Read more

Multivariate normal distribution is normalized (of course): A proof

-
Image
Let \(\mathbf{X} = (X_1, X_2, \cdots, X_n)^\top\) be a vector of random variables. We say it follows the multivariate normal (Gaussian) distribution if its density is given by \[f(\mathbf{x}) = \frac{1}{\sqrt{(2\pi)^n|\Sigma|}}\exp\left(-\frac{1}{2}(\mathbf{x} - \boldsymbol{\mu})^\top\Sigma^{-1}(\mathbf{x} - \boldsymbol{\mu})\right)\tag{Eq:density}\] where \(\boldsymbol{\mu} = (\mu_1, \mu_2, \cdots, \mu_n)^\top \in \mathbb{R}^n\) is a vector, \(\Sigma\) is a symmetric   positive definite \(n\times n\) matrix, and \(\Sigma^{-1}\) and \(|\Sigma|\) are the inverse and determinant of \(\Sigma\), respectively. It turns out that \(\boldsymbol{\mu}\) and \(\Sigma\) are the mean vector and covariance matrix of \(\mathbf{X}\), respectively. But we will not prove that here. In this post, we will show this density (Eq:density) is normalized (of course). That is, we prove that \[\int_{\mathbb{R}^n}f(\mathbf{x})d\mathbf{x} = 1.\] We assume that you already know how to prove the univariate nor...
Read more