Hans Andersen Brendekilde (1857-1942) was a Danish painter.
We have sections
English Video.
(Part 1 of Limits) (Part 2 of Limits) (Indeterminate Form for Limits of Functions) (Continuity of Functions)
中文影音.
(函數極限 1) (函數極限 2) (不定型之函數極限) (函數的連續性)
\begin{equation}{\label{a}}\tag{A}\mbox{}\end{equation}
The Idea and Definition of Limit.
Let \(f:\mathbb{R}\rightarrow\mathbb{R}\) be a real-valued function. The limit
\[\lim_{x\rightarrow c} f(x)=L\]
can be interpreted as follows
\[\mbox{if \(x\) approaches \(c\) then \(f(x)\) approaches \(L\)}.\]
Now, we are going to consider the right-hand limit and left-hand limit. The right- and left-hand limit are also called one-sided limit.
The left-hand limit
\[\lim_{x\rightarrow c-} f(x)=L\mbox{ means that if \(x\) approaches \(c\) from the left, then \(f(x)\) approaches \(L\)}.\]
The right-hand limit
\[\lim_{x\rightarrow c+} f(x)=L\mbox{ means that if \(x\) approaches \(c\) from the right, then \(f(x)\) approaches \(L\)}.\]
We can show that
\[\lim_{x\rightarrow c} f(x)=L\mbox{ if and only if }\lim_{x\rightarrow c-} f(x)=L=\lim_{x\rightarrow c+} f(x).\]
Suppose that
\[\lim_{x\rightarrow c-} f(x)\neq\lim_{x\rightarrow c+} f(x).\]
Then, the limit
\[\lim_{x\rightarrow c} f(x)\]
does not exist.
Example. Let \(f\) be a function defined by
\[f(x)=\frac{|x|}{x}\mbox{ for }x\neq 0.\]
This function can be written as
\[f(x)=\left\{\begin{array}{ll}
1 & \mbox{if \(x>0\)}\\
-1 & \mbox{if \(x<0\)}
\end{array}\right .\]
We first consider the limit
\[\lim_{x\rightarrow 0} f(x)\]
We have
\[\lim_{x\rightarrow 0-} f(x)=\lim_{x\rightarrow 0-} (-1)=-1\]
and
\[\lim_{x\rightarrow 0+} f(x)=\lim_{x\rightarrow 0+} (1)=1.\]
Therefore, the limit
\[\lim_{x\rightarrow 0}\frac{|x|}{x}\]
does not exist. Suppose that \(c_{1}>0\) and \(c_{2}<0\). Then
\[\lim_{x\rightarrow c_{1}}\frac{|x|}{x}=1\mbox{ and }\lim_{x\rightarrow c_{2}}\frac{|x|}{x}=-1\]
Example. Consider the following limit
\[\lim_{x\rightarrow 2} \frac{1}{x-2}\]
The function \(f(x)=1/(x-2)\) is not defined at \(x=2\). We also have
\[\lim_{x\rightarrow 2+}\frac{1}{x-2}=+\infty\mbox{ and }\lim_{x\rightarrow 2-}\frac{1}{x-2}=-\infty .\]
Example. Consider the limit
\[\lim_{x\rightarrow 4}\frac{x^{2}-2x-8}{x-4}=\lim_{x\rightarrow 4} (x+2)=6.\]
Example. Consider the following function
\[f(x)=\left\{\begin{array}{ll}
3x-1 & \mbox{if \(x\neq 2\)}\\
45 & \mbox{if \(x=2\)}
\end{array}\right .\]
Then we have
\[\lim_{x\rightarrow 2} f(x)=\lim_{x\rightarrow 2-} f(x)=\lim_{x\rightarrow 2+} f(x)=5.\]
Example. Consider the following function
\[f(x)=\left\{\begin{array}{ll}
1-x^{2} & \mbox{if \(x<1\)}\\
2x & \mbox{if \(x>1\)}
\end{array}\right .\]
Then, the limit
\[\lim_{x\rightarrow 1} f(x)\]
does not exist, since the one-sided limits
\[\lim_{x\rightarrow 1-} f(x)=\lim_{x\rightarrow 1-} (1-x^{2})=0\mbox{ and }\lim_{x\rightarrow 1+} f(x)=\lim_{x\rightarrow 1+} 2x=2\]
are different
In the sequel, we are going to provide the formal definition of limit. Let \(f:\mathbb{R}\rightarrow\mathbb{R}\) be a real-valued function, and let \(c\) be a real number. The function \(f\) is not necessarily defined at \(c\). However, we need to assume that the function \(f\) is defined on a set of the form
\[(c-p,c)\cup (c,c+p)\mbox{ with }p>0.\]
This consideration will guarantee that we can consider \(f(x)\) for all \(x\neq c\) that are “sufficiently close” to \(c\).
Definition. We consider a real-valued function \(f:\mathbb{R}\rightarrow\mathbb{R}\).
- (The Limit of a Function). Let \(f\) be defined on a set of the following form
\[(c-p,c)\cup (c,c+p)\mbox{ with }p>0.\]
Then, we write
\[\lim_{x\rightarrow c}f(x)=L\]
when, for each \(\varepsilon >0\), there exists \(\delta >0\) such that
\[\mbox{$0<|x-c|<\delta$ implies \(|f(x)-L|<\epsilon\).}\] - (Left-Hand Limit of a Function). Let \(f\) be defined on an interval of the form \((c-p,c)\) with \(p>0\).
Then, we write
\[\lim_{x\rightarrow c-} f(x)=L\]
when, for each \(\varepsilon >0\), there exists \(\delta >0\) such that
\[\mbox{$c-\delta<x<c$ implies \(|f(x)-L|<\varepsilon\).}\] - (Right-Hand Limit of a Function). Let \(f\) be defined on an interval of the form \((c,c+p)\) with \(p>0\).
Then, we write
\[\lim_{x\rightarrow c+} f(x)=L\]
when, for each \(\varepsilon >0\), there exists \(\delta >0\) such that
\[\mbox{$c<x<c+\delta$ implies \(|f(x)-L|<\varepsilon\).}\]
As shown above, the one-sided limits can provide us a technique to determine whether or not a (two-sided) limit exists. This is given below.
\[\lim_{x\rightarrow c} f(x)=L\mbox{ if and only if }\lim_{x\rightarrow c-} f(x)=L\mbox{ and }\lim_{x\rightarrow c+} f(x)=L.\]
Example. Show that
\[\lim_{x\rightarrow 2} (2x-1)=3.\]
Let \(\varepsilon >0\). By definition, we are going to find a \(\delta >0\) such that
\[\mbox{$0<|x-2|<\delta$ implies \(|(2x-1)-3|<\varepsilon\).}\]
From \(|(2x-1)-3|<\varepsilon\), we have \(|x-2|<\varepsilon /2\). This says that we can take \(\delta =\varepsilon /2\). Therefore, we have that
\[\mbox{$0<|x-2|<\varepsilon /2$ implies \(2|x-2|<\varepsilon\).}\]
In other words, we obtain \(|(2x-1)-3|<\varepsilon\). \(\sharp\)
Example. Find the limit
\[\lim_{x\rightarrow c}|x|.\]
By definition, if we can prove that, for each \(\varepsilon >0\), there exists \(\delta >0\) such that \(0<|x-c|<\delta\) implies \(||x|-|c||<\varepsilon\), then the limit is given by
\[\lim_{x\rightarrow c} |x|=|c|.\]
Since \(||x|-|c||\leq |x-c|\), (recall \(||a|-|b||\leq |a-b|\)), we can take \(\delta =\varepsilon\). This shows that
\[0<|x-c|<\varepsilon\mbox{ implies }||x|-|c||<\varepsilon.\]
Therefore, we obtain the desired result. \(\sharp\)
The following limits are equivalent:
- \({\displaystyle \lim_{x\rightarrow c} f(x)=L;}\)
- \({\displaystyle \lim_{h\rightarrow 0}f(x+h)=L;}\)
- \({\displaystyle \lim_{x\rightarrow c} (f(x)-L)=0;}\)
- \({\displaystyle \lim_{x\rightarrow c} |f(x)-L|=0.}\)
\begin{equation}{\label{b}}\tag{B}\mbox{}\end{equation}
Some Limit Theorems.
Theorem. (The Uniqueness of Limit). Suppose that
\[\lim_{x\rightarrow c} f(x)=L\mbox{ and }\lim_{x\rightarrow c} f(x)=M.\]
Then, we have \(L=M\). \(\sharp\)
Theorem. Suppose that
\[\lim_{x\rightarrow c} f(x)=L\mbox{ and }\lim_{x\rightarrow c} g(x)=M.\]
Then, we have the following properties
\begin{align*}
& \lim_{x\rightarrow c} [f(x)+g(x)]=L+M=\lim_{x\rightarrow c} f(x)+\lim_{x\rightarrow c} g(x)\\
& \lim_{x\rightarrow c} [\alpha f(x)]=\alpha L=\alpha\lim_{x\rightarrow c} f(x)\mbox{ for all real \(\alpha\)}\\
& \lim_{x\rightarrow c} [f(x)g(x)]=LM=\left [\lim_{x\rightarrow c} f(x)\right ]\left [\lim_{x\rightarrow c} g(x)\right ].
\end{align*}
Note that
\[f(x)-g(x)=f(x)+(-1)g(x).\]
Therefore, we have
\[\lim_{x\rightarrow c} [f(x)-g(x)]=L-M=\lim_{x\rightarrow c} f(x)-\lim_{x\rightarrow c} g(x).\]
Suppose that
\[\lim_{x\rightarrow c} f_{1}(x)=L_{1},\quad\lim_{x\rightarrow c} f_{2}(x)=L_{2},\quad\cdots ,\quad\lim_{x\rightarrow c} f_{n}(x)=L_{n}.\]
Then, we have
\[\lim_{x\rightarrow c} [\alpha_{1}f_{1}(x)+\alpha_{2}f_{2}(x)+\cdots +\alpha_{n}f_{n}(x)]=\alpha_{1}L_{1}+\alpha_{2}L_{2}+\cdots +\alpha_{n}L_{n}\]
and
\[\lim_{x\rightarrow c} [f_{1}(x)f_{2}(x)\cdots f_{n}(x)]=L_{1}L_{2}\cdots L_{n}.\]
Let
\[P(x)=a_{n}x^{n}+\cdots +a_{1}x+a_{0}\]
be a polynomial, and let \(c\) be any real number. Then, we have
\[\lim_{x\rightarrow c} P(x)=P(c).\]
Theorem. We have the following properties.
(i) Suppose that
\[\lim_{x\rightarrow c} g(x)=M\mbox{ with }M\neq 0.\]
Then, we have
\[\lim_{x\rightarrow c}\frac{1}{g(x)}=\frac{1}{M}.\]
(ii) Suppose that
\[\lim_{x\rightarrow c} f(x)=L\mbox{ and }\lim_{x\rightarrow c} g(x)=M\mbox{ with }M\neq 0.\]
Then, we have
\[\lim_{x\rightarrow c} \frac{f(x)}{g(x)}=\frac{L}{M}.\]
(iii) Suppose that
\[\lim_{x\rightarrow c} f(x)=L\mbox{ with }L\neq 0\mbox{ and }\lim_{x\rightarrow c} g(x)=0.\]
Then, the limit
\[\lim_{x\rightarrow c} \frac{f(x)}{g(x)}\]
does not exist. \(\sharp\)
Example. Evaluate the following limit
\[\lim_{x\rightarrow 2} \frac{1/x-1/2}{x-2}.\]
For \(x\neq 2\), we have
\[\frac{1/x-1/2}{x-2}=\frac{(2-x)/2x}{x-2}=\frac{-1}{2x}.\]
Therefore, we obtain
\[\lim_{x\rightarrow 2} \frac{1/x-1/2}{x-2}=-\frac{1}{4}.\]
Example. Evaluate the following limit
\[\lim_{x\rightarrow 9} \frac{x-9}{\sqrt{x}-3}.\]
First of all, we rationalize the denominator
\[\frac{x-9}{\sqrt{x}-3}=\frac{x-9}{\sqrt{x}-3}\cdot\frac{\sqrt{x}+3}{\sqrt{x}+3}=\sqrt{x}+3.\]
Then, we have the limit
\[\lim_{x\rightarrow 9} \frac{x-9}{\sqrt{x}-3}=\lim_{x\rightarrow 9}[\sqrt{x}+3]=6.\]
Example. Evaluate the following limit
\[\lim_{x\rightarrow 9}\left (\frac{x-9}{\sqrt{x}-3}+\frac{1/x-1/9}{x-9}\right ).\]
Since
\[\lim_{x\rightarrow 9}\frac{1/x-1/9}{x-9}=\lim_{x\rightarrow 9} \frac{9-x}{(9x)(x-9)}=-\frac{1}{81},\]
the limit is
\[\lim_{x\rightarrow 9}\left (\frac{x-9}{\sqrt{x}-3}+\frac{1/x-1/9}{x-9}\right )=6-\frac{1}{81}=\frac{485}{81}.\]
Example. Evaluate the following limit
\[\lim_{x\rightarrow 9} \frac{1/x-1/9}{\sqrt{x}-3}.\]
We have
\begin{align*} \lim_{x\rightarrow 9} \frac{1/x-1/9}{\sqrt{x}-3} & =\lim_{x\rightarrow 9}
\frac{x-9}{\sqrt{x}-3}\cdot\frac{1/x-1/9}{x-9}\\ & =6\cdot\left (-\frac{1}{81}\right )=-\frac{2}{27}.\end{align*}
\begin{equation}{\label{c}}\tag{C}\mbox{}\end{equation}
Continuity.
Definition. Let \(f\) be a function defined at least on an open interval
\[(c-p,c+p)\mbox{ with }p>0.\]
Then \(f\) is continuous at \(c\) when
\[\lim_{x\rightarrow c} f(x)=f(c).\]
The function \(f\) is said to be discontinuous at \(c\) if it is not continuous at \(c\). \(\sharp\)
According to the definition of limit, we see that \(f\) is continuous at \(c\) if and only if, for each \(\varepsilon >0\), there exists \(\delta >0\) such that
\[|x-c|<\delta\mbox{ implies }|f(x)-f(c)|<\varepsilon.\]
A function \(f\) is continuous at \(c\) if and only if the following conditions are satisfied:
- \(f\) is defined at \(c\);
- \({\displaystyle \lim_{x\rightarrow c} f(x)}\) exists;
- \({\displaystyle \lim_{x\rightarrow c} f(x)=f(c)}\).
Theorem. Suppose that \(f\) and \(g\) are continuous at \(c\). Then, we have the following properties.
(i) The addition \(f+g\) is continuous at \(c\).
(ii) The scalar multiplication \(\alpha f\) is continuous at \(c\).
(iii) The product \(fg\) is continuous at \(c\).
(iv) The quotient \(f/g\) is continuoius at \(c\) if \(g(c)\neq 0\). \(\sharp\)
Theorem. Suppose that \(g\) is continuous at \(c\) and \(f\) is continuous at \(g(c)\). Then, the composition \(f(g(x))\) is continuous at \(c\). \(\sharp\)
Definition. (One-Sided Continuity). A function \(f\) is called continuous at \(c\) from the left when
\[\lim_{x\rightarrow c-} f(x)=f(c).\]
A function \(f\) is called continuous at \(c\) from the right when
\[\lim_{x\rightarrow c+} f(x)=f(c). \sharp\]
It is clear to see that a function \(f\) is continuous at \(c\) if and only if
\[f(c)=\lim_{x\rightarrow c-} f(x)=\lim_{x\rightarrow c+} f(x).\]
Example. Determine the discontinuities of the following function
\[f(x)=\left\{\begin{array}{ll}
2x+1 & \mbox{if \(x\leq 0\)}\\
1 & \mbox{if \(0<x\leq 1\)}\\
x^{2}+1 & \mbox{if \(x>1\)}
\end{array}\right .\]
Clearly, \(f\) is continuous at each point in the open interval \((-\infty ,0)\), \((0,1)\) and \((1,+\infty )\). Now, we have
\begin{align*} & \lim_{x\rightarrow 0-} f(x)=\lim_{x\rightarrow 0-} (2x+1)=1,\\
& \lim_{x\rightarrow 0+} f(x)=\lim_{x\rightarrow 0+} 1\\ & f(0)=1,\end{align*}
which say that \(f\) is continuous at \(0\). Since
\[\lim_{x\rightarrow 1-} f(x) =\lim_{x\rightarrow 1-} 1=1\]
and
\[\lim_{x\rightarrow 1+} f(x)=\lim_{x\rightarrow 1+} (x^{2}+1)=2,\]
it says that \(f\) is discontinuous at \(1\). \(\sharp\)
Example. Determine the discontinuities of the following function
\[f(x)=\left\{\begin{array}{ll}
x^{3}, & x\leq -1\\
x^{2}-2, & -1<x<0\\
3-x, & 0\leq x<2\\
{\displaystyle \frac{4x-1}{x-1}}, & 2\leq x<4\\ &\\ {\displaystyle \frac{15}{7-x}}, & 4<x<7\\ &\\
5x+2, & 7\leq x.
\end{array}\right .\]
It is clear that \(f\) is continuous at each point in the open intervals \((-\infty ,-1)\), \((-1,0)\), \((0,2)\), \((2,4)\), \((4,7)\) and \((7,\infty )\). We have to check the behavior of \(f\) at \(x=-1,0,2,4\) and \(7\). The results are summarized in the following table
\[\begin{array}{|c|c|c|c|c|}\hline
c & f(c) & {\displaystyle \lim_{x\rightarrow c-} f(x)} & {\displaystyle \lim_{x\rightarrow c+} f(x)} & \mbox{Conclusion}\\ \hline
-1 & -1 & -1 & -1 & \mbox{Continuous}\\ \hline
0 & 3 & -2 & 3 & \mbox{Discontinuous}\\ \hline
2 & 7 & 1 & 7 & \mbox{Discontinuous}\\ \hline
4 & \mbox{Not defined} & 5 & 5 & \mbox{Discontinuous}\\ \hline
7 & 37 & \mbox{Does not exist} & 37 & \mbox{Discontinuous}\\ \hline
\end{array}\]
\begin{equation}{\label{d}}\tag{D}\mbox{}\end{equation}
The Pinching Theorem.
Theorem. (The Pinching Theorem). Suppose that \(h(x)\leq f(x)\leq g(x)\) satisfy
\[\lim_{x\rightarrow c} h(x)=L\mbox{ and }\lim_{x\rightarrow c} g(x)=L.\]
Then, we have
\[\lim_{x\rightarrow c} f(x)=L.\]
Example. Find the limit
\[\lim_{x\rightarrow 0}\frac{\sin x}{x}.\]
Comparing the areas, we have the following inequality
\[\frac{1}{2}\sin x\leq\frac{1}{2}x\leq\frac{1}{2}\frac{\sin x}{\cos x}.\]
For \(x>0\), we obtain
\[\cos x\leq\frac{\sin x}{x}\leq 1.\]
Since \(\cos (-x)=\cos x\) and
\[\frac{\sin (-x)}{-x}=\frac{-\sin x}{-x}=\frac{\sin x}{x},\]
the above inequality also holds for \(x<0\). We can obtain
\[\lim_{x\rightarrow 0} \frac{\sin x}{x}=1\]
by applying the pinching theorem. \(\sharp\)
\begin{equation}{\label{e2}}\tag{1}\mbox{}\end{equation}
Example \ref{e2}. Find the limit
\[\lim_{x\rightarrow 0} \frac{1-\cos x}{x}.\]
We have
\begin{align*}
\frac{1-\cos x}{x} & =\left (\frac{1-\cos x}{x}\right )\left (\frac{1+\cos x}{1+\cos x}\right )\\
& =\frac{1-\cos^{2}x}{x(1+\cos x)}\\
& =\frac{\sin^{2}x}{x(1+\cos x)}\\
& =\left (\frac{\sin x}{x}\right )\left (\frac{\sin x}{1+\cos x}\right ).
\end{align*}
Since
\[\lim_{x\rightarrow 0} \frac{\sin x}{x}=1\mbox{ and }\lim_{x\rightarrow 0} \frac{\sin x}{1+\cos x}=0,\]
it follows
\[\lim_{x\rightarrow 0} \frac{1-\cos x}{x}=0.\]
Example. Find the limit
\[\lim_{x\rightarrow 0} \frac{\sin 4x}{x}.\]
Since
\[\lim_{x\rightarrow 0}\frac{\sin x}{x}=1,\]
it follows
\[\lim_{x\rightarrow 0}\frac{\sin 4x}{4x}=1.\]
We obtain
\[\lim_{x\rightarrow 0} \frac{\sin 4x}{x}=4.\]
Example. Find the limit
\[\lim_{x\rightarrow 0} x\cot 3x.\]
We write
\[x\cot 3x=x\frac{\cos 3x}{\sin 3x}=\frac{1}{3}\left (\frac{3x}{\sin 3x}\right )(\cos 3x).\]
Since
\[\lim_{x\rightarrow 0}\frac{\sin x}{x}=1\mbox{ is equivalent to }\lim_{x\rightarrow 0}\frac{x}{\sin x}=1,\]
we obtain
\begin{align*} \lim_{x\rightarrow 0} x\cot 3x & =\frac{1}{3}\left (\lim_{x\rightarrow 0}
\frac{3x}{\sin 3x}\right )\left (\lim_{x\rightarrow 0} \cos 3x\right )\\ & =
\frac{1}{3}\cdot 1\cdot 1=\frac{1}{3}.\end{align*}
Example. Find the limit
\[\lim_{x\rightarrow \pi /4}\frac{\sin (x-\frac{\pi}{4})}{(x-\pi/4)^{2}}.\]
We write
\[\frac{\sin (x-\pi /4)}{(x-\pi /4)^{2}}=\left [\frac{\sin (x-\pi /4)}{x-\pi /4}\right ]\cdot\frac{1}{x-\pi /4}\]
Let \(y=x-\pi /4\). Then \(x\rightarrow\pi /4\) implies \(y\rightarrow 0\). We have
\[\lim_{x\rightarrow\pi /4} \frac{\sin (x-\pi /4)}{x-\pi /4}=
\lim_{y\rightarrow 0}\frac{\sin y}{y}=1.\]
Since
\[\lim_{x\rightarrow \pi /4} (x-\pi /4)=0,\]
the limit
\[\lim_{x\rightarrow \pi /4} \frac{\sin (x-\pi /4)}{(x-\pi /4)^{2}}\]
does not exist. \(\sharp\)
Example. Find the limit
\[\lim_{x\rightarrow 0}\frac{x^{2}}{\sec x-1}.\]
We have
\begin{align*}
\frac{x^{2}}{\sec x-1} & =\frac{x^{2}}{\sec x-1}\left (\frac{\sec x+1}{\sec x+1}\right )\\
& =\frac{x^{2}(\sec x+1)}{\sec^{2} x-1}\\
& =\frac{x^{2}(\sec x+1)}{\tan^{2} x}\\
& =\frac{x^{2}\cos^{2} x(\sec x+1)}{\sin^{2} x}\\
& =\left (\frac{x}{\sin x}\right )^{2}(\cos^{2} x)(\sec x+1).
\end{align*}
Therefore, we obtain the limit
\begin{align*} \lim_{x\rightarrow 0}\frac{x^{2}}{\sec x-1} & =\lim_{x\rightarrow 0}\left (\frac{x}{\sin x}\right )^{2}\cdot
\lim_{x\rightarrow 0}\cos^{2} x\cdot\lim_{x\rightarrow 0} (\sec x+1)\\ & =1\cdot 1\cdot 2=2.\end{align*}
Example. Find the limit
\[\lim_{x\rightarrow\pi /4}\frac{\sin\left (x+\frac{\pi}{4}\right )-1}{x-\frac{\pi}{4}}.\]
We have
\begin{align*} \sin\left (x+\frac{\pi}{4}\right ) & =\sin\left (x-\frac{\pi}{4}+\frac{\pi}{2}\right )\\ & =\cos\left (x-\frac{\pi}{4}\right ).\end{align*}
Let \(y=x-\pi /4\). Then, \(x\rightarrow\pi /4\) implies \(y\rightarrow 0\). Therefore, using Example \ref{e2}, we obtain
\begin{align*}
\lim_{x\rightarrow\pi /4}\frac{\sin\left (x+\frac{\pi}{4}\right )-1}{x-\frac{\pi}{4}} & =\lim_{x\rightarrow\pi /4}\frac{\cos\left (x-\frac{\pi}{4}\right )-1}
{x-\frac{\pi}{4}}\\
& =\lim_{y\rightarrow 0}\frac{\cos y-1}{y}\\ & =-\lim_{y\rightarrow 0}\frac{1-\cos y}{y}=0.
\end{align*}
