Learning Outcomes
- Compute limits of functions given algebraically, using proper limit properties.
Section 1.3 Limits analytically (LT3)
Subsection 1.3.1 Activities
Remark 1.3.1.
Recall that in Activity 1.2.5 we used numerical methods and table of values to find the limit of a relatively simple degree three polynomial at a point. This was inefficient, “there’s gotta be a better way!”
Activity 1.3.2.
Given \(f(x)=3x^2-\frac{1}{2}x+4\text{,}\) evaluate \(f(2)\) and approximate \(\displaystyle \lim_{x\to2}f(x)\) numerically (or graphically). What do you think is more likely?
- \(\displaystyle \displaystyle \lim_{x \to 2}f(x)=f(2)\)
- \(\displaystyle \displaystyle \lim_{x \to 2}f(x) \approx f(2)\)
- \(\displaystyle \displaystyle \lim_{x \to 2}f(x) \neq f(2)\)
Activity 1.3.3.
The table below gives values of a few different functions.
| x | 6.99 | 6.999 | 7.001 | 7.01 |
| f(x) | 13.99 | 13.999 | 14.001 | 14.01 |
| g(x) | 22.97 | 22.997 | 23.003 | 23.03 |
| 3f(x) | 41.97 | 41.997 | 42.003 | 42.03 |
| f(x)+g(x) | 36.96 | 36.996 | 37.004 | 37.04 |
| f(x)g(x) | 321.350 | 321.935 | 322.065 | 322.650 |
Using the table above, which of the following is least likely to be true?
- \(\displaystyle \lim_{x\to 7}f(x)= 14\) and \(\displaystyle \lim_{x\to 7}g(x)= 23\)
- \(\displaystyle \displaystyle \lim_{x \to 7}3f(x) = 3 \lim_{x\to 7}f(x)\)
- \(\displaystyle \displaystyle \lim_{x\to 7}\left( f(x)+g(x) \right) = \lim_{x\to 7}f(x) + \lim_{x\to 7}g(x)\)
- \(\displaystyle \displaystyle \lim_{x\to 7}\left( f(x)g(x) \right) =f(7) \left( \lim_{x\to 7}g(x) \right)\)
Remark 1.3.4.
In Activity 1.3.3 we observed that limits seem to be "well-behaved" when combined with standard operations on functions. The next theorems, known as Limit Laws, tell us how limits interact with combinations of functions.
Theorem 1.3.5. Limit Laws, I.
Let \(f\) and \(g\) be functions defined on an open interval \(I\) containing the real number \(c\) satisfying
\begin{equation*}
\lim_{x \to c} f(x) = L \, \text{and} \, \lim_{x \to c} g(x) = K,
\end{equation*}
for \(L\) and \(K\) some real numbers. Then we have the following limits.
- Constant Law: \(\displaystyle \lim_{x \to c} b = b\text{,}\) for \(b\) any constant real number;
- Identity Law: \(\displaystyle \lim_{x \to c} x = c\) ;
- Sum/Difference Law: \(\displaystyle \lim_{x \to c} (f(x) \pm g(x)) = L \pm K\text{;}\)
- Scalar Multiple Law: \(\displaystyle \lim_{x \to c} b \cdot f(x) = bL\text{,}\) for \(b\) any constant real number;
- Product Law: \(\displaystyle \lim_{x \to c} f(x) \cdot g(x) = LK\text{;}\)
- Quotient Law: if \(K \ne 0\text{,}\) then \(\displaystyle \lim_{x \to c} f(x) / g(x) = L/K\text{.}\)
Activity 1.3.6.
If \(\displaystyle \lim_{x\to 2} f(x) = 2\) and \(\displaystyle \lim_{x\to 2} g(x) = -3\text{,}\) which of the following statements are true? Select all that apply!
- \(\displaystyle \displaystyle \lim_{x\to 2} (f(x) \cdot g(x)) = -6\)
- \(\displaystyle \displaystyle \lim_{x\to 2} (f(x) + g(x)) = -1\)
- \(\displaystyle \displaystyle \lim_{x\to 2} (f(x) - g(x)) = -2\)
- \(\displaystyle \displaystyle \lim_{x\to 2} (f(x)/g(x)) = -2/3\)
Theorem 1.3.7. Limit Laws, II.
Let \(f\) and \(g\) be functions defined on an open interval \(I\) containing \(c\) satisfying
\begin{equation*}
\lim_{x \to c} f(x) = L, \, \, \, \, \lim_{x \to L} g(x) = K, \text{and} \, g(L)=K.
\end{equation*}
Then we have the following limits as well.
- Power Law: \(\displaystyle \lim_{x \to c} f(x)^n = L^n\text{,}\) for \(n\) a positive integer;
- Root Law: \(\displaystyle \lim_{x \to c} \sqrt[n]{f(x)} = \sqrt[n]{L}\text{,}\) for \(n\) a positive integer;
- Composition Law: \(\displaystyle \lim_{x \to c} g(f(x)) = K. \)
Activity 1.3.8.
Below you are given the graphs of two functions. Compute the limits below (if possible).
(a)
\(\displaystyle \lim_{x \to 1} f(x) + g(x) \text{.}\)
(b)
\(\displaystyle \lim_{x \to 5^+} 3f(x) \text{.}\)
(c)
\(\displaystyle \lim_{x \to 0^+ } f(x)g(x) \text{.}\)
(d)
(Challenge) \(\displaystyle \lim_{x \to 1} g(x) / f(x) \text{.}\)
(e)
(Challenge) \(\displaystyle \lim_{x \to 0^+} f(g(x)) \text{.}\)
Activity 1.3.9.
Given \(p(x) = -3x^2 - 5x+7\text{,}\) which of the following limit laws would use to determine \(\displaystyle \lim_{x\to 2} p(x)\text{?}\) Choose all that apply.
- Sums/Difference Law
- Scalar Multiple Law
- Product Law
- Identity Law
- Power Law
- Constant Law
Theorem 1.3.10. Limits of Polynomials.
If \(p(x)\) is a polynomial and \(c\) is a real number, then \(\displaystyle \lim_{x \to c} p(x) = p(c)\text{.}\) This is also known as the Direct Substitution Property for polynomials.
Activity 1.3.11.
Given \(p(x)=-3x^2-5x+7\) and \(q(x)=x^4-x^2+3\text{,}\) which of the following describes the most efficient way to determine \(\displaystyle \lim_{x \to -1} \frac{p(x)}{q(x)}\text{?}\)
- Sums/difference, scalar multiple, and product laws
- Theorem 1.3.10 and the quotient law
- Power, sums/difference, scalar multiple, and constant laws
- Quotient and root law
Theorem 1.3.12. Limits of Rational Functions.
If \(p(x)\) and \(q(x)\) are polynomials, \(c\) is a real number, and \(q(c) \neq 0\) then \(\displaystyle \lim_{x \to c} \frac{p(x)}{q(x)}=\frac{p(c)}{q(c)}\text{.}\)
Activity 1.3.13.
Consider taking the limit of a rational function \(\frac{p(x)}{q(x)}\) as \(x \to c\text{.}\) If \(q(c)=0\text{,}\) is it possible for \(\displaystyle\lim_{x \to c}\frac{p(x)}{q(x)} \) to equal a number?
- No, because \(\frac{p(x)}{q(x)}\) is not defined at \(x=c\) since \(q(c)=0\text{.}\)
- Yes, because if you graph \(f(x)=\frac{x^2-1}{x-1}\text{,}\) the value \(f(1)\) is not defined, but the graph shows that the limit of \(f(x)\) does exist as \(x \to 1\text{.}\)
- No, because if you graph \(g(x)=\frac{x^2+1}{x-1}\text{,}\) the value \(g(1)\) is not defined and the graph shows that the limit of \(\displaystyle\lim_{x \to c}g(x)\) does not exist.
- Yes, because we can use Theorem 1.3.12.
Activity 1.3.14.
Let \(f(x) = 2x\) and \(g(x) = x\text{,}\) which of the following statements is true?
- \(\displaystyle \displaystyle \lim_{x\to 0} (f(x)/g(x)) = 0\)
- \(\displaystyle \displaystyle \lim_{x\to 0} (f(x)/g(x)) = 2\)
- \(\displaystyle \lim_{x\to 0} (f(x)/g(x))\) cannot be determined
- \(\displaystyle \lim_{x\to 0} (f(x)/g(x))\) does not exist
Remark 1.3.15.
When we compute the limit of a ratio where both the numerator and denominator have limit equal to zero, we have to compute the value of a \(\frac{0}{0}\) indeterminate form. The value of an indeteminate form can be any real number or even infinity or not existent, we just do not know yet! We can usually determine the value of an indeterminate form using some algebraic manipulations of the expression given.
Definition 1.3.16.
A function \(f(x)\) has a hole at \(x=c\) if \(f(c)\) does not exist but \(\displaystyle \lim_{x \to c} f(x)\) does exist and is equal to a real number.
Example 1.3.17.
The function \(f(x)=\frac{x^2-1}{x-1}\) has a hole at \(x=1\) because \(f(1)\) is not defined but
\begin{equation*}
\lim_{x \to 1} \frac{x^2-1}{x-1} = \lim_{x \to 1} \frac{(x-1)(x+1)}{x-1} = \lim_{x \to 1} (x+1) = 2 ,
\end{equation*}
so the limit exists and is equal to a real number. Notice that \(\displaystyle \lim_{x \to 1} \frac{x^2-1}{x-1}\) is also an example of a limit giving an indeterminate form \(\frac{0}{0}\) which we could then compute using an algebraic manipulation of the function given.
Activity 1.3.18.
Determine the following limits and explain your reasoning.
(a)
\begin{equation*}
\lim_{x\to-6 } \frac{ x^{2} - 6 \, x + 5 }{ x^{2} - 3 \, x - 18 }
\end{equation*}
(b)
\begin{equation*}
\lim_{x\to-1 } \frac{ x^{2} - 1 }{ x^{2} + 3 \, x + 2 }
\end{equation*}
(c)
\begin{equation*}
\lim_{x\to5 } \frac{ x - 5 }{ \sqrt{x + 31} - 6 }
\end{equation*}
Activity 1.3.19.
In activity Activity 1.2.8 you studied the velocity of Usain Bolt in his Beijing 100 meters dash. We will now study this situation analytically. To make our computations simpler, we will approximate that he could run 100 meters in 10 seconds and we will consider the model \(d=f(t)=t^2\text{,}\) where \(d\) is the distance in meters and \(t\) is the time in seconds.
Note 1.3.20.
The average velocity is the ratio distance covered over time elapsed. If we consider the interval that starts at \(t=a\) and has width \(h\text{,}\) written \([a,a+h]\text{,}\) the average velocity on this interval is \(\displaystyle \frac{f(a+h)-f(a)}{(a+h) - a} = \frac{f(a+h)-f(a)}{h}\text{.}\) The instantaneous velocity at time \(t=a\) is given by:
\begin{equation*}
\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}\text{.}
\end{equation*}
(a)
Compute the average velocity on the interval \([5,6]\text{.}\) We think of this interval as \([5,5+h]\) for the value of \(h=1\text{.}\)
(b)
Compute the average velocity starting at 5 seconds, but now with \(h=0.5\) seconds.
(c)
We want to study the instantaneous velocity at \(a=5\) seconds. Find an expression for the average velocity on the interval \([5,5+h]\text{,}\) where \(h\) is an unspecified value.
(d)
Expand your expression. When \(h \neq 0\text{,}\) you can simplify it!
(e)
Recall that the instantaneous velocity is the limit of your expression as \(h\to 0\text{.}\) Find the instantaneous velocity given by this model at \(t=5\) seconds.
(f)
The model \(d=f(t)=t^2\) does not really capture the real-world situation. Think of at least one reason why this model does not fit the scenario of Usain Bolt’s 100 meters dash.
