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Workbook for Stewart’s "Calculus"

Cory Wilson

Section 1.6 Calculating Limits Using Limit Laws

Subsection 1.6.1 Before Class

https://mymedia.ou.edu/media/1.6-1/1_1uboropy
Figure 8. Pre-Class Video 1

Subsubsection 1.6.1.1 Limit Laws

  1. Sum/Difference Rule: The limit of a sum/difference of functions is the sum/difference of the limit of the functions.
  2. Constant Multiplier Rule: The limit of a constant times a function is the constant times the limit of the function.
  3. Product Rule: The limit of a product of functions is the product of the limits.
  4. Quotient Rule: The limit of a quotient of functions is the quotient of the limits (provided the denominator is nonzero).
  5. Power Rule: The limit of a function raised to a power is the power of the limit of the function.
  6. Constant Rule: The limit of a constant is a constant.
  7. Root Rule: The limit of the \(n\)th-root of a function is the \(n\)th-root of the limit of a function.
Example 1.6.2.
Use the limit laws to determine the following limits, if they exist. The function \(f(x)\) is graphed as the solid line, and \(g(x)\) is graphed as the dashed line.
The function \(f(x)\) is graphed as the solid line, and \(g(x)\) is graphed as the dashed line.
  1. \(\displaystyle \ds \lim_{x\to 1} [f(x)]\)
  2. \(\displaystyle \ds \lim_{x\to -1} [g(x)]\)
  3. \(\displaystyle \ds \lim_{x\to -2}\left[\dfrac{f(x)}{g(x)}\right]\)
  4. \(\displaystyle \ds \lim_{x\to -1}\left[f(x)g(x)\right]\)
  5. \(\displaystyle \ds \lim_{x\to -1}\left[-2f(x)+3g(x)\right]\)
  6. \(\displaystyle \ds \lim_{x\to 2} [3f(x)]\)
Solution.
  1. \(\displaystyle \ds \lim_{x\to 1} [f(x)] \text{ DNE}\)
  2. \(\displaystyle \ds \lim_{x\to -1} [g(x)]=1\)
  3. \(\displaystyle \ds \lim_{x\to -2}\left[\dfrac{f(x)}{g(x)}\right]=-\dfrac{2}{3}\)
  4. \(\displaystyle \ds \lim_{x\to -1}\left[f(x)g(x)\right]= -1\)
  5. \(\displaystyle \ds \lim_{x\to -1}\left[-2f(x)+3g(x)\right]=5\)
  6. \(\displaystyle \ds \lim_{x\to 2} [3f(x)]=0\)
Direct Substitution Property.
If \(f\) is a polynomial, rational, or algebraic function with \(a\) in the domain of \(f\text{,}\) then \(\ds \lim_{x\to a} f(x) = f(a)\)
Example 1.6.3.
Calculate the following limits. Justify each answer with one (or more) of the limit laws.
  1. \(\displaystyle \ds \lim_{x\to -1} 3x^3+5x^2-7\)
  2. \(\displaystyle \ds \lim_{x\to 2} \dfrac{x^3+2x^2-1}{5-3x}\)
  3. \(\displaystyle \ds \lim_{x\to 1} \dfrac{x-1}{x^2+1}\)
Solution.
  1. \(\ds \lim_{x\to -1} 3x^3+5x^2-7 = -9\text{.}\) Since this is a polynomial, you can use the direct substitution property.
  2. \(\ds \lim_{x\to 2} \dfrac{x^3+2x^2-1}{5-3x}=-15\text{.}\) Since this is a rational function with 2 in the domain, we can use direct substitution.
  3. \(\ds \lim_{x\to 1} \dfrac{x-1}{x^2+1}=0\text{.}\) Since this is a rational function with 1 in the domain, we can use direct substitution.
Question 1.6.4.
Let \(k(x)=\dfrac{x+1}{x^2-1}\text{.}\)
  1. The Direct Substitution Property cannot be applied to the function at \(x = -1\text{.}\) Why not?
  2. Modify the function so that you can use the Direct Substitution Property.
Solution.
  1. \(-1\) is not in the domain
  2. Factor the denominator as \((x+1)(x-1)\) and cancel to get \(\dfrac{1}{x-1}\text{.}\)

Subsection 1.6.2 Pre-Class Activities

Example 1.6.5.

Assume that \(\ds \lim_{x\to 3} f(x) = -1\text{,}\) \(\ds \lim_{x\to 3} g(x) = 2\text{,}\) and \(\ds \lim_{x\to 3} h(x) = 0\text{.}\) Find the following limits, if they exist. If they don’t, explain why. Justify each answer with the appropriate limit law(s).
  1. \(\displaystyle \ds \lim_{x\to 3} [f(x) + 5g(x)]\)
  2. \(\displaystyle \ds \lim_{x\to 3} [f(x)]^5\)
  3. \(\displaystyle \ds \lim_{x\to 3} \sqrt{g(x)}\)
  4. \(\displaystyle \ds \lim_{x\to 3} \dfrac{3f(x)}{g(x)}\)
  5. \(\displaystyle \ds \lim_{x\to 3} \dfrac{g(x)}{h(x)}\)
  6. \(\displaystyle \ds \lim_{x\to 3} \dfrac{f(x)h(x)}{g(x)}\)
Solution.
  1. \(\ds \lim_{x\to 3} [f(x) + 5g(x)] = 9\text{.}\) Use the sum/difference law to split the limit, then use the constant multiple rule to multiply \(\ds \lim_{x\to 3} g(x)\) by 5.
  2. \(\ds \lim_{x\to 3} [f(x)]^5=-1\text{.}\) Use the power rule to take the power out of the limit, then evaluate.
  3. \(\ds \lim_{x\to 3} \sqrt{g(x)}=\sqrt{2}\text{.}\) Use the root rule to take the root out of the limit, then evaluate.
  4. \(\ds \lim_{x\to 3} \dfrac{3f(x)}{g(x)}=-\dfrac{3}{2}\text{.}\) Use the quotient rule to split into two limits, then the constant multiple rule to get \(\ds \lim_{x\to 3} f(x)\) by itself.
  5. \(\ds \lim_{x\to 3} \dfrac{g(x)}{h(x)}\) does not exist. There is a non-removable factor of 0 in the denominator.
  6. \(\ds \lim_{x\to 3} \dfrac{f(x)h(x)}{g(x)}=0\text{.}\) Use the quotient rule to split into two limits, then the product rule to split the numerator into two limits as well.

Subsection 1.6.3 In Class

Example 1.6.6.

Compute the following limits:
  1. \(\ds \lim_{t\to -5} g(t)\text{,}\) where \(g(t) = \begin{cases}t^2 -1 \amp t\neq -5\\ \pi+1 \amp t = -5 \end{cases}\)
  2. \(\ds \lim_{h\to 0} \dfrac{f(x+h)-f(x)}{h}\text{,}\) where \(f(x) = x^2+2x-1\)
  3. \(\ds \lim_{h\to 0} \dfrac{f(1+h)-f(1)}{h}\text{,}\) where \(f(x) = \sqrt{x}\)
Solution.
  1. \(\displaystyle \ds \lim_{t\to -5} g(t) = 24\)
  2. \(\displaystyle \ds \lim_{h\to 0} \dfrac{f(x+h)-f(x)}{h} = 2x+2\)
  3. \(\displaystyle \ds \lim_{h\to 0} \dfrac{f(1+h)-f(1)}{h}=\dfrac{1}{2}\)

Example 1.6.7.

Show that \(\ds \lim_{x\to a} |x-a| = 0\)
Solution.
Since we can’t use the direct substitution property, we split into right- and left-hand limits.
\begin{equation*} \ds \lim_{x\to a^+} |x-a| = \lim_{x\to a^+} (x-a) = a-a = 0 \end{equation*}
\begin{equation*} \ds \lim_{x\to a^-} |x-a| = \lim_{x\to a^-} -(x-a) = -a+a = 0 \end{equation*}
Since the left- and right-hand limits are the same, we conclude that
\begin{equation*} \ds \lim_{x\to a} |x-a|=0 \end{equation*}

Definition 1.6.8. Greatest Integer Function.

The greatest integer function, \([\![x]\!]\text{,}\) is the largest integer that is less than or equal to \(x\text{.}\)

Example 1.6.9.

Find \([\![x]\!]\) for the following values of \(x\text{.}\)
  1. \(\displaystyle x = 2\)
  2. \(\displaystyle x = -1\)
  3. \(\displaystyle x = 5.1\)
  4. \(\displaystyle x = \pi\)
  5. \(\displaystyle x = -6.7\)
Solution.
  1. \(\displaystyle 2\)
  2. \(\displaystyle -1\)
  3. \(\displaystyle 5\)
  4. \(\displaystyle 3\)
  5. \(\displaystyle -7\)

Example 1.6.10.

Find the limits.
  1. \(\displaystyle \ds\lim_{x\to 0.5} [\![x]\!]\)
  2. \(\displaystyle \ds\lim_{x\to 2} [\![x]\!]\)
  3. \(\displaystyle \ds \lim_{x\to 3} \dfrac{\frac{1}{x}-\frac{1}{3}}{x-3}\)
  4. \(\displaystyle \ds\lim_{h\to 1} \dfrac{h^4-1}{h^3-1}\)
Solution.
  1. \(\displaystyle \ds\lim_{x\to 0.5} [\![x]\!] = 0\)
  2. \(\displaystyle \ds\lim_{x\to 2} [\![x]\!] \text{ DNE}\)
  3. \(\displaystyle \ds \lim_{x\to 3} \dfrac{\frac{1}{x}-\frac{1}{3}}{x-3} = -\dfrac{1}{9}\)
  4. \(\displaystyle \ds\lim_{h\to 1} \dfrac{h^4-1}{h^3-1} = \dfrac{4}{3}\)

Subsubsection 1.6.3.1 Two Theorems

Example 1.6.13.
Use the Squeeze Theorem to show that \(\ds \lim_{x\to 0} x^2 \sin \lrpar{\dfrac{1}{x}}= 0\)
Solution.
Since \(-1\leq \sin\theta \leq 1\text{,}\) it must be true that \(-1\leq \sin\lrpar{\dfrac{1}{x}}\leq 1\text{.}\) Then, \(-x^2\leq x^2\sin\lrpar{\dfrac{1}{x}}\leq x^2\text{.}\) Since \(\ds\lim_{x\to 0} (-x^2) = 0\) and \(\ds \lim_{x\to 0} x^2 = 0\text{,}\) we conclude that \(\ds \lim_{x\to 0} x^2 \sin \lrpar{\dfrac{1}{x}}= 0\) by the Squeeze Theorem.

Subsection 1.6.4 After Class Activities

Example 1.6.14.

If \(2x\leq g(x) \leq x^6 - x^4 + 2\) for all \(x\text{,}\) find \(\ds \lim_{x\to 1} g(x)\text{.}\)
Solution.
\(\ds \lim_{x\to 1} g(x) = 2\) by the Squeeze Theorem

Example 1.6.15.

The signum function (or the sign function), \(\text{sgn}(x)\text{,}\) is given by
\begin{equation*} \text{sgn}(x) = \begin{cases}-1 \amp x \lt 0\\ 0 \amp x = 0\\ 1 \amp x > 0 \end{cases} \end{equation*}
  1. Sketch the graph of \(\text{sgn}(x)\text{.}\)
  2. Find the limits, or explain why they don’t exist:
    1. \(\displaystyle \ds \lim_{x\to 0} \text{sgn}(x)\)
    2. \(\displaystyle \ds \lim_{x\to 0} |\text{sgn}(x)|\)
Solution.
  1. The graph of the signum function on the interval \([-2,2]\text{.}\)
  2. Find the limits, or explain why they don’t exist:
    1. \(\ds \lim_{x\to 0} \text{sgn}(x) \text{ DNE}\text{.}\) This is because the left and right hand limits disagree.
    2. \(\displaystyle \ds \lim_{x\to 0} |\text{sgn}(x)| = 1\)

Example 1.6.16.

Find the limit:
  1. \(\displaystyle \ds \lim_{x\to -4} \dfrac{3x + 12}{|x + 4|}\)
  2. \(\displaystyle \ds \lim_{x\to 2} \dfrac{\sqrt{4x+1} - 3}{x-2}\)
  3. \(\displaystyle \ds \lim_{x\to -1} \dfrac{2x^2 + 3x + 1}{x^2 - 2x -3}\)
  4. \(\displaystyle \ds \lim_{h\to 0} \dfrac{(x+h)^3 - x^3}{h}\)
Solution.
  1. \(\displaystyle \ds \lim_{x\to -4} \dfrac{3x + 12}{|x + 4|} \text{ DNE}\)
  2. \(\displaystyle \ds \lim_{x\to 2} \dfrac{\sqrt{4x+1} - 3}{x-2} = \dfrac{2}{3}\)
  3. \(\displaystyle \ds \lim_{x\to -1} \dfrac{2x^2 + 3x + 1}{x^2 - 2x -3} = \dfrac{1}{4}\)
  4. \(\displaystyle \ds \lim_{h\to 0} \dfrac{(x+h)^3 - x^3}{h} = 3x^2\)
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