for all real numbers at \(x=2\) for all real numbers, except \(x=2\) impossible to say
We give basic laws for working with limits.
In the previous section were able to compute the limits
\[ \lim _{x\to \pi } x^3=\pi ^3,\quad \lim _{x\to \pi } \sqrt {2}=\sqrt {2},\quad \lim _{x\to \pi } \cos {x}= -1, \]
using continuity of the functions \(x^3\), \(\sqrt {2}\), and \(\cos {x}\), at \(x=\pi \). Does this imply that we can compute the limits \begin{align*} &\lim _{x\to \pi } (x^3-\cos {x}),\\ &\lim _{x\to \pi } \frac {\sqrt {2}}{\cos {x}},\\ &\lim _{x\to \pi } (\sqrt {2}\cdot x^3\cdot \cos {x}), \text { or}\\ &\lim _{x\to \pi } \cos ({x^3})? \end{align*}
Well, we cannot use continuity here, because we don’t know if the functions \(x^3-\cos {x}\), \(\frac {\sqrt {2}}{\cos {x}}\), \(\sqrt {2}\cdot x^3\cdot \cos {x}\), and \(\cos ({x^3})\) are continuous at \(x=\pi \), and
we have no other tools available, since the graphs and tables are not reliable. Obviously, we need more tools to help us with computation of limits.
In this section, we present a handful of rules, called the Limit Laws, that allow us to find limits of various combinations of functions.
Limit Laws Suppose that \(\lim _{x\to a}f(x)=L\), \(\lim _{x\to a}g(x)=M\).
- Sum/Difference Law
-
\(\lim _{x\to a} (f(x) \pm g(x)) = \lim _{x\to a}f(x) \pm \lim _{x\to a}g(x)=L \pm M\).
- Product Law
-
\(\lim _{x\to a} (f(x)g(x)) = \lim _{x\to a}f(x)\cdot \lim _{x\to a}g(x)=LM\).
- Quotient Law
-
\(\lim _{x\to a} \frac {f(x)}{g(x)} = \frac {\lim _{x\to a}f(x)}{\lim _{x\to a}g(x)}=\frac {L}{M}\), if \(M\ne 0\).
In plain language, “limit of a sum equals the sum of the limits,” “limit of a product equals the product of the limits,” etc.
Let’s examine how the Limit Laws can be used in computation of limits.
Compute the following limits using Limit Laws:
- (a)
- \(\lim _{x\to \pi } (x^3-\cos {x})\)
- (b)
- \(\lim _{x\to \pi } \frac {\sqrt {2}}{\cos {x}}\)
- (c)
- \(\lim _{x\to \pi } (\sqrt {2}\cdot x^3\cdot \cos {x})\)
- (d)
- \(\lim _{x\to \pi } \cos ({x^3})\)
For \(\lim _{x\to \pi } (x^3-\cos {x})\), write: \begin{align*} \lim _{x\to \pi } (x^3-\cos {x})&=\lim _{x\to \pi } x^3-\lim _{x\to \pi }\cos {x} && \text {by Difference Law}\\ &=\pi ^3-(-1)\\ &=\pi ^3+1 \end{align*}
For \(\lim _{x\to \pi } \frac {\sqrt {2}}{\cos {x}}\), write: \begin{align*} \lim _{x\to \pi } \frac {\sqrt {2}}{\cos {x}}&=\frac {\lim _{x\to \pi }\sqrt {2}}{\lim _{x\to \pi }\cos {x}} && \text {by Quotient Law}\\ &=\frac {\sqrt {2}}{-1}\\ &=-\sqrt {2} \end{align*}
For \(\lim _{x\to \pi } (\sqrt {2}\cdot x^3\cdot \cos {x})\), using the Product Law, we can write:
\[ \lim _{x\to \pi } (\sqrt {2}\cdot x^3\cdot \cos {x})=\lim _{x\to \pi } \sqrt {2}\cdot \lim _{x\to \pi }x^3\cdot \lim _{x\to \pi }\cos {x} \]
\begin{align*} &=\sqrt {2}\cdot \pi ^3\cdot (-1)\\ &=-\sqrt {2}\pi ^3 \end{align*}
For \(\lim _{x\to \pi } \cos ({x^3})\), write:
\[ \lim _{x\to \pi } \cos ({x^3})= \cdots \]
The function \(\cos ({x^3})\) is a combination of the functions \(\cos ({x})\) and \(x^3\), but it is neither a
sum/difference, nor a product, nor a quotient of these two functions, so we cannot
apply any of the Limit Laws. This function is a composition of the two functions, \(\cos ({x})\)
and \(x^3\). - (a)
- Compute the following limit using limit laws: \[ \lim _{x\to 1}(5x^2+3x-2) \]Well, get out your pencil and write with me:\[ \lim _{x\to 1} (5x^2+3x-2) = \lim _{x\to 1} 5x^2 + \lim _{x\to 1} \answer [given]{3x} - \lim _{x\to 1}2 \]by the Sum/Difference Law. So now\[ = 5\lim _{x\to 1} x^2 + 3\lim _{x\to 1} x - \lim _{x\to 1}\answer [given]{2} \]by the Product Law. Finally by continuity of \(x^b\) and \(k\),\[ = 5(1)^2 + 3(1) - 2 =\answer [given]{6}. \]We can check our answer by looking at the graph of \(y=f(x)\):\[ \graph {5x^2+3x-2} \]
- (b)
- Is the polynomial function \(f(x)=5x^2+3x-2\) continuous at \(x=1\)? We have to check whether\[ \lim _{x\to 1} f(x)=f(1). \]Since we already know that \( \lim _{x\to 1} (5x^2+3x-2)=6\), we only have to compute \(f(1)\). \begin{align*} f(1)&=5(1)^2 + 3(1) - 2\\ &=6 \end{align*}
Therefore, the polynomial function \(f\) is continuous at \(x=1\).
But what about continuity at any other value \(x\)? Is the function \(f\) continuous on its entire domain?
And what about any other polynomial function?
Are polynomials continuous on their domains?
We can generalize the example above to get the following theorems.
Continuity of Polynomial Functions All polynomial functions, meaning
functions of the form
\[ f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1 x + a_0 \]
where \(n\) is a positive integer and each coefficient \(a_i\), \(i=0, 1,...,n\), is
a real number, are continuous for all real numbers. In order to show
that any polynomial, \(f\), is continuous at any real number, \(a\), we have to show
that
\[ \lim _{x\to a} f(x)=f(a). \]
Write with me: \[ \lim _{x\to a} f(x) = \lim _{x\to a} (a_nx^n + a_{n-1}x^{n-1} + \dots + a_1 x + a_0 ) \]
Now by the Sum Law, \[ = \lim _{x\to a} a_nx^n + \lim _{x\to a} a_{n-1}x^{n-1} + \dots + \lim _{x\to a}a_1 x + \lim _{x\to a} a_0 \]
and by the Product Law,
\begin{align*} = \lim _{x\to a} a_n\cdot \lim _{x\to a}x^n + \lim _{x\to a} a_{n-1}\cdot \lim _{x\to a}x^{n-1} &+ \dots \\ &+ \lim _{x\to a}a_1 \cdot \lim _{x\to a}x + \lim _{x\to a} a_0 \end{align*}
and by Continuity
\[ = a_n\cdot a^n + a_{n-1}\cdot a^{n-1} + \dots + a_1 \cdot a + a_0 \]
And this equals…\[ =f(a) \]
Since we have shown that \(\lim _{x\to a} f(x) = f(a)\), we have shown that \(f\)
is continuous at \(x=a\). Continuity of Rational Functions A rational function h, meaning a function of the
form
\[ h(x)=\frac {f(x)}{g(x)} \]
where \(f \)and \(g\) are polynomials, is continuous for all real numbers except where \(g(x)=0\).
That is, rational functions are continuous wherever they are defined. Let \(a\) be a real
number such that \(g(a)\neq 0\). Then, since \(g(x)\) is continuous at \(a\), \(\lim _{x\to a} g(x) \neq 0\). Therefore, write with me,
\[ \lim _{x \to a} h(x) = \lim _{x\to a} \frac {f(x)}{g(x)} \]
and
now by the Quotient Law, \[ \frac {\lim _{x\to a} f(x)}{ \lim _{x\to a} g(x)} \]
and by the continuity of polynomials we may now
set \(x=a\) \[ \frac {f(a)}{g(a)}=h(a). \]
Since we have shown that \(\lim _{x\to a} h(x) = h(a)\), we have shown that \(h\) is continuous at \(x=a\). Where is \(f(x) = \frac {x^2-3x+2}{x-2}\) continuous?
True or false: If \(f\) and \(g\) are continuous functions on an interval \(I\),
then \(f\pm g\) is continuous on \(I\).
True False
Let’s assume that \(I\) is an open interval and \(a\)
is a number in \(I\). Remember, since \(f\) and \(g\) are both continuous on \(I\), they are both
continuous at \(a\).
This means that \( \lim _{x\to a}f(x)=f(a)\) and \( \lim _{x\to a}g(x)=g(a)\).
Now, define a new function, \(h\), where \(h(x)= f(x)+g(x)\), for all \(x\) in \(I\). We have to show that \(h\) is continuous at \(a\), or that
\[ \lim _{x\to a} h(x) = h(a). \]
Lat’s start with
\[ \lim _{x\to a} h(x) = \lim _{x\to a} (f(x)+g(x)) = \lim _{x\to a} f(x)+\lim _{x\to a}g(x)\hspace {0.04in}by \hspace {0.04in}Sum\hspace {0.04in} Law, \]
and, therefore,
\[ \lim _{x\to a} h(x) = f(a)+g(a)=h(a) \checkmark \]
We have proved that \(h\) is continuous at any number \(a\) in \(I\). Therefore, \(h\) is continuous on \(I\).
Similarly, we can prove that \(f+g\) is continuous on any interval \(I\), by showing it is left-or
right-continuous at the endpoints. We can adjust the proof for the function \(f-g\). True or false: If \(f\) and \(g\) are continuous functions on an interval \(I\), then \(f/g\) is continuous
on \(I\).
True False
In this case, \(f/g\) will not be continuous for \(x\) where \(g(x) = 0\).
We still don’t know how to compute a limit of a composition of two functions. Our next theorem provides basic rules for how limits interact with composition of functions.
Composition Limit Law If \(f(x)\) is continuous at \(b = \lim _{x\to a} g(x)\), then
\[ \lim _{x\to a} f(g(x)) = f(\lim _{x\to a} g(x)). \]
Because the limit of a continuous function is the same as the function value, we can now pass limits inside continuous functions.
Continuity of Composite Functions
If \(g\) is continuous at \(x=a\), and if \(f \) is continuous at \(g(a)\), then \(f(g(x))\) is continuous at \(x=a\).
Using the Composition Limit Law, we can compute the last example from the beginning of this section.
Compute the following limit using limit laws:
\[ \lim _{x \to \pi }\cos ({x^3}) \]
We will use
the Composition Limit Law. Let
\[ f(g(x))=\cos ({x^3}), \]
where \(f(x)=\cos {x}\), and \(g(x)=x^3\). Now, continuity of \(x^3\) implies that \(\lim _{x \to \pi }g(x)=\pi ^3\).
Continuity of \(\cos {x}\) implies that \(f\) is continuous at \(\pi ^3=\lim _{x \to \pi }g(x)\). Now, the Composition Limit Law
implies that \[ \lim _{x \to \pi }\cos ({x^3})= \cos \Bigl ({\lim _{x \to \pi } x^3}\Bigr )=\cos ({\pi ^3})\approx 0.91726. \]
We can confirm our results by checking out the graph of \(y=f(g(x))\):
\[ \graph {\cos ({x^3})} \]
Many of the Limit Laws and theorems about continuity in this section might seem like they should be obvious. You may be wondering why we spent an entire section on these theorems. The answer is that these theorems will tell you exactly when it is easy to find the value of a limit, and exactly what to do in those cases.
The most important thing to learn from this section is whether the limit laws can be applied for a certain problem, and when we need to do something more interesting. We will begin discussing those more interesting cases in the next section.
1 A list of questions
Let’s try this out.
Can this limit be directly computed by limit laws?
\[ \lim _{x\to 2}\frac {x^2+3x+2}{x+2} \]
yes no
Compute:
\[ \lim _{x\to 2}\frac {x^2+3x+2}{x+2}\begin{prompt} =\answer {3}\end{prompt} \]
Since \(f(x)=\frac {x^2+3x+2}{x+2}\) is a rational function, and the denominator does not equal \(0\), we see
that \(f(x)\) is continuous at \(x=2\). Thus, to find this limit, it suffices to plug \(2\) into \(f(x)\).
Can this limit be directly computed by limit laws?
\[ \lim _{x\to 2}\frac {x^2-3x+2}{x-2} \]
yes no
\(f(x) = \frac {x^2-3x+2}{x-2}\) is a rational
function, but the denominator \(x-2\) equals \(0\) when \(x=2\). None of our current theorems address
the situation when the denominator of a fraction approaches \(0\).
Can this limit be directly computed by limit laws?
\[ \lim _{x\to 0} x\sin (1/x) \]
yes no
If we are trying to
use limit laws to compute this limit, we would first have to use the Product Law to
say that
\[ \lim _{x\to 0}x\sin (1/x)= \lim _{x\to 0} x \cdot \lim _{x\to 0} \sin (1/x). \]
We are only allowed to use this law if both limits exist, so we must check
this first. We know from continuity that \[ \lim _{x\to 0}x=0. \]
However, we also know that \(\sin (1/x)\) oscillates
“wildly” as \(x\) approaches \(0\), and so the limit \[ \lim _{x\to 0} \sin (1/x) \]
does not exist. Therefore, we cannot use the
Product Law. Can this limit be directly computed by limit laws?
\[ \lim _{x\to 0} \cot (x^3) \]
yes no
Notice that
\[ \cot (x^3) = \frac {\cos (x^3)}{\sin (x^3)}. \]
If we
are trying to use limit laws to compute this limit, we would like to use the Quotient
Law to say that \[ \lim _{x\to 0} \frac {\cos (x^3)}{\sin (x^3)} = \frac {\lim _{x\to 0} \cos (x^3)}{\lim _{x\to 0} \sin (x^3)}. \]
We are only allowed to use this law if both limits exist and the
denominator is not \(0\). We suspect that the limit in the denominator might equal \(0\), so
we check this limit. \begin{align*} \lim _{x\to 0} \sin (x^3) &= \sin (\lim _{x\to 0}x^3)\\ &=\sin (0) \\ &=0. \end{align*}
This means that the denominator is zero and hence we cannot use the Quotient Law.
Can this limit be directly computed by limit laws?
\[ \lim _{x\to 0}\sec ^2(e^x-1) \]
yes no
Compute:
\[ \lim _{x\to 0}\sec ^2(e^x-1)\begin{prompt} =\answer {1}\end{prompt} \]
Notice
that
\[ \lim _{x\to 0} \sec ^2(e^x-1) = \lim _{x\to 0} \frac {1}{\cos ^2(e^x-1)}. \]
If we are trying to use Limit Laws to compute this limit, we would now have to
use the Quotient Law to say that \[ \lim _{x\to 0} \frac {1}{\cos ^2(e^x-1)} = \frac { \lim _{x\to 0}1}{ \lim _{x\to 0}\cos ^2(e^x-1)}. \]
We are only allowed to use this law if
both limits exist and the denominator is not \(0\). Let’s check the denominator
and numerator separately. First we’ll compute the limit of the denominator:
\begin{align*} \lim _{x\to 0}\cos ^2(e^x-1) &= (\lim _{x\to 0}\cos {(e^x-1)})^{2}\\ &= \Bigl (\cos {\Big (\lim _{x\to 0}(e^x-1)\Bigr )}\Bigr )^{2}\\ & =\cos ^2\Bigl (\lim _{x\to 0}(e^x-1)\Bigr )\\ &=\cos ^2\Bigl (\lim _{x\to 0}(e^x)-\lim _{x\to 0}(1)\Bigr )\\ &=\cos ^2(1-1)\\ &= \cos ^2(0)\\ &=1 \end{align*}
Therefore, the limit in the denominator exists and does not equal \(0\). We can use the Quotient Law, so we will compute the limit of the numerator:
\[ \lim _{x\to 0}1=1 \]
Hence \[ \frac { \lim _{x\to 0}1}{ \lim _{x\to 0}\cos ^2(e^x-1)} = \frac {1}{1}=1 \]
Can this limit be directly computed by limit laws?
\[ \lim _{x\to 1}{(x-1)\cdot \csc (\ln (x))} \]
yes no
If we are trying to
use limit laws to compute this limit, we would have to use the Product Law to say
that
\[ \lim _{x\to 1}{(x-1)\cdot \csc (\ln (x))}= \lim _{x\to 1}{(x-1)\cdot \lim _{x\to 1}\csc (\ln (x))}. \]
We are only allowed to use this law if both limits exist. Let’s check each limit
separately. \begin{align*} \lim _{x\to 1} (x-1) &= \lim _{x\to 1} (x)-\lim _{x\to 1}(1)\\ &=1-1\\ &=0. \end{align*}
So this limit exists. Now we check the other factor. Notice that
\[ \lim _{x\to 1}\csc (\ln (x)) = \frac {1}{\sin (\ln (x))}. \]
If we are trying to
use limit laws to compute this limit, we would now have to use the Quotient Law to
say that \[ \frac {1}{\sin (\ln (x))} = \frac {\lim _{x\to 1}1}{\lim _{x\to 1}\sin (\ln (x))}. \]
We are only allowed to use this law if both limits exist and the denominator
does not equal \(0\). The limit in the numerator definitely exists, so let’s check the limit
in the denominator. \begin{align*} \lim _{x\to 1}\sin (\ln (x)) &= \sin (\lim _{x\to 1}\ln (x))\\ &=\sin (\ln (1))\\ &= \sin (0)\\ &= 0 \end{align*}
Since the denominator is \(0\), we cannot apply the Quotient Law.
Can this limit be directly computed by limit laws?
\[ \lim _{x\to 0} x\ln x \]
yes no
If we are trying to
use limit laws to compute this limit, we would have to use the Product Law to say
that
\[ \lim _{x\to 0} x\ln x =\lim _{x\to 0} x \cdot \lim _{x\to 0}\ln x. \]
We are only allowed to use this law if both limits exist. We know \(\lim _{x\to 0} x = 0\), but what
about \(\lim _{x\to 0}\ln x\)? We do not know how to find \(\lim _{x\to 0}\ln x\) using limit laws because \(0\) is not in the domain
of \(\ln x\). Can this limit be directly computed by limit laws?
\[ \lim _{x\to 0}\frac {2^x-1}{3^{x-1}} \]
yes no
Compute:
\[ \lim _{x\to 0}\frac {2^x-1}{3^{x-1}}\begin{prompt} =\answer {0}\end{prompt} \]
If we
are trying to use limit laws to compute this limit, we would have to use the Quotient
Law to say that
\[ \lim _{x\to 0}\frac {2^x-1}{3^{x-1}} = \frac {\lim _{x\to 0}(2^x-1)}{\lim _{x\to 0}(3^{x-1})}. \]
We are only allowed to use this law if both limits exist and the
denominator does not equal \(0\). Let’s check each limit separately, starting with the
denominator \begin{align*} \lim _{x\to 0}(3^{x-1}) &=3^{\lim _{x\to 0}(x-1)}\\ &=3^{-1}\\ &=\frac {1}{3} \end{align*}
On the other hand the limit in the numerator is \begin{align*} \lim _{x\to 0}(2^x-1) &=\lim _{x\to 0}(2^x)-\lim _{x\to 0}(1)\\ &=1-1\\ &=0 \end{align*}
The limits in both the numerator and denominator exist and the limit in the denominator does not equal \(0\), so we can use the Quotient Law. We find:
\[ \frac {\lim _{x\to 0}(2^x-1)}{\lim _{x\to 0}(3^{x-1})} =\frac {0}{\frac {1}{3}}=0. \]
Can this limit be directly computed by limit laws?
2025-01-06 19:46:35 \[ \lim _{x\to 0}(1+x)^{1/x} \]
yes no
We do not have
any limit laws for functions of the form \(f(x)^{g(x)}\), so we cannot compute this limit.