1.4.3: Additional Exercises

Reading Questions

1. What is the "Mathematical Mantra" introduced in this section?
2. According to the theorem on locally equivalent functions, if \(f(x) = g(x)\) for all \(x \neq a\) in an open interval containing \(a\), what can be said about \( \displaystyle \lim_{x \to a} f(x)\) and \( \displaystyle \lim_{x \to a} g(x)\)?
3. If direct substitution into \( \displaystyle \lim_{x \to a} \frac{N(x)}{D(x)}\) results in the indeterminate form \(\frac{0}{0}\), what general strategy does this section suggest?
4. Name three algebraic techniques discussed in this section for manipulating a function to evaluate a limit when direct substitution fails.
5. When evaluating a limit by factoring and canceling, why is it permissible to cancel the common factor even though it makes the original function undefined at that point?
6. When using conjugate multiplication to evaluate a limit, which parts of the resulting expression is it generally advisable to distribute, and which parts to leave undistributed? Why?
7. What is the piecewise definition of the absolute value function \(|u|\)?
8. How does the piecewise definition of the absolute value function help in evaluating limits like \( \displaystyle \lim_{x \to (-5/3)^-} \frac{|3x+5|}{6x^2+7x-5}\)?
9. In Example 5, \( \displaystyle \lim_{x \to 0} \left( \frac{1}{x} + \frac{5}{x(x-5)} \right)\), why can't the Sum Law for limits be applied directly? What algebraic step is performed first?
10. When working with limits, until what point in the calculation should the "lim" notation be written?

Homework

In exercises 1 - 10, use the Direct Substitution Property to show that each limit leads to the indeterminate form \(0/0\). Then, evaluate the limit analytically.

1) \(\displaystyle \lim_{x \to 4}\frac{x^2−16}{x−4}\)

Answer

\(\displaystyle \text{When }x = 4, \quad\frac{x^2−16}{x−4}=\frac{16−16}{4−4}=\frac{0}{0};\)

then, \(\displaystyle \lim_{x \to 4}\frac{x^2−16}{x−4}= \lim_{x \to 4}\frac{(x+4)(x−4)}{x−4}=\lim_{x \to 4}(x+4) = 4+4 =8\)

2) \(\displaystyle \lim_{x \to 2}\frac{x−2}{x^2−2x}\)

3) \(\displaystyle \lim_{x \to 6}\frac{3x−18}{2x−12}\)

Answer

\(\displaystyle \text{When }x = 6, \quad\frac{3x−18}{2x−12}=\frac{18−18}{12−12}=\frac{0}{0};\)

then, \(\displaystyle \lim_{x \to 6}\frac{3x−18}{2x− 12}=\lim_{x \to 6}\frac{3(x−6)}{2(x−6)}=\lim_{x \to 6}\frac{3}{2}=\frac{3}{2}\)

4) \(\displaystyle \lim_{h \to 0}\frac{(1+h)^2−1}{h}\)

5) \(\displaystyle \lim_{t \to 9}\frac{t−9}{\sqrt{t}−3}\)

Answer

\(\displaystyle \text{When }t = 9, \quad\frac{t−9}{\sqrt{t}−3}=\frac{9−9}{3−3}=\frac{0}{0};\)

then, \(\displaystyle \lim_{t \to 9}\frac{t−9}{\sqrt{t}−3} =\lim_{t \to 9}\frac{t−9}{\sqrt{t}−3}\frac{\sqrt{t}+3}{\sqrt{t}+3}=\lim_{t \to 9}\frac{(t−9)(\sqrt{t}+3)}{t - 9}=\lim_{t \to 9}(\sqrt{t}+3)=\sqrt{9}+3=6\)

6) \(\displaystyle \lim_{h \to 0}\frac{\dfrac{1}{a+h}−\dfrac{1}{a}}{h}\), where \(a\) is a real-valued constant

7) \(\displaystyle \lim_{ \theta \to \pi }\frac{\sin \theta }{\tan \theta }\)

Answer

\(\displaystyle \text{When } \theta = \pi , \quad\frac{\sin \theta }{\tan \theta }=\frac{\sin \pi }{\tan \pi }=\frac{0}{0};\)

then, \(\displaystyle \lim_{ \theta \to \pi }\frac{\sin \theta }{\tan \theta }=\lim_{ \theta \to \pi }\frac{\sin \theta }{\frac{\sin \theta }{\cos \theta }}=\lim_{ \theta \to \pi }\cos \theta =\cos \pi =−1\)

8) \(\displaystyle \lim_{x \to 1}\frac{x^3−1}{x^2−1}\)

9) \(\displaystyle \lim_{x \to 1/2}\frac{2x^2+3x−2}{2x−1}\)

Answer

\(\displaystyle \text{When }x=1/2, \quad\frac{2x^2+3x−2}{2x−1}=\frac{\frac{1}{2}+\frac{3}{2}−2}{1−1}=\frac{0}{0};\)

then, \(\displaystyle \lim_{x \to 1/2}\frac{2x^2+3x−2}{2x−1}=\lim_{x \to 1/2}\frac{(2x−1)(x+2)}{2x−1}=\lim_{x \to 1/2}(x+2)=\frac{1}{2}+2=\frac{5}{2}\)

10) \(\displaystyle \lim_{x \to −3}\frac{\sqrt{x+4}−1}{x+3}\)

In exercises 11 - 14, use the Direct Substitution Property to obtain an undefined expression, simplify the function, and determine the limit.

11) \(\displaystyle \lim_{x \to −2^−}\frac{2x^2+7x−4}{x^2+x−2}\)

Answer

\(− \infty \)

12) \(\displaystyle \lim_{x \to −2^+}\frac{2x^2+7x−4}{x^2+x−2}\)

13) \(\displaystyle \lim_{x \to 1^−}\frac{2x^2+7x−4}{x^2+x−2}\)

Answer

\(− \infty \)

14) \(\displaystyle \lim_{x \to 1^+}\frac{2x^2+7x−4}{x^2+x−2}\)