Chapter 2, Problem 1RE

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

1. a. The rational function $\frac{x-1}{x^2-1}$ has vertical asymptotes at x = −1 and x = 1.
2. b. Numerical or graphical methods always produce good estimates of $\underset{x\to a}{\lim }f(x)$.
3. c. The value of $\underset{x\to a}{\lim }f(x)$, if it exists, is found by calculating f(a).
4. d. If $\underset{x\to a}{\lim }f(x)=\infty$ or $\underset{x\to a}{\lim }f(x)=-\infty$, then $\underset{x\to a}{\lim }f(x)$ does not exist.
5. e. If $\underset{x\to a}{\lim }f(x)$ does not exist, then either $\underset{x\to a}{\lim }f(x)=\infty$ or $\underset{x\to a}{\lim }f(x)=-\infty$.
6. f. The line y = 2x + 1 is a slant asymptote of the function $f(x)=\frac{2x^2+x}{x-3}$.
7. g. If a function is continuous on the intervals (a, b) and [b, c), where a < b < c, then the function is also continuous on (a, c).
8. h. If $\underset{x\to a}{\lim }f(x)$ can be calculated by direct substitution, then f is continuous at x = a.

To determine

Whether the statement “The rational function $\frac{x-1}{x^2-1}$ has vertical asymptotes at $x=-1$ and $x=1$” is true or false.

Answer to Problem 1RE

The statement is false.

Explanation of Solution

Consider the rational function $f\left(x\right)=\frac{x-1}{x^2-1}$ (1)

Take limit in the equation (1) as x approaches 1.

$\begin{array}{c}\underset{x\to 1}{\lim }f\left(x\right)=\underset{x\to 1}{\lim }\frac{x-1}{x^2-1}\\ =\underset{x\to 1}{\lim }\frac{\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\underset{x\to 1}{\lim }\frac{1}{\left(x+1\right)}\\ =\frac{1}{2}\end{array}$

Therefore, the line $x=1$ is not a vertical asymptote.

Take limit in the equation (1) as x approaches −1.

$\begin{array}{c}\underset{x\to -1}{\lim }f\left(x\right)=\underset{x\to -1}{\lim }\frac{x-1}{x^2-1}\\ =\underset{x\to -1}{\lim }\frac{\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\underset{x\to -1}{\lim }\frac{1}{\left(x+1\right)}\\ =\frac{1}{-1+1}\end{array}$

It can be simplified as $\underset{x\to -1}{\lim }f\left(x\right)=\infty$.

Therefore, the line $x=-1$ is a vertical asymptote.

Therefore, the given statement is false.

To determine

Whether the statement “Numerical or graphical methods always produce good estimate of $\underset{x\to a}{\lim }f\left(x\right)$” is true or false.

Answer to Problem 1RE

The statement is false.

Explanation of Solution

The numerical and graphical methods does not produce accurate value because the value of x is near to a but $x\ne a$.

Therefore, the given statement is false.

To determine

Whether the statement “The value of $\underset{x\to a}{\lim }f\left(x\right)$, if it exists, is found by calculating $f\left(a\right)$” is true or false.

Answer to Problem 1RE

The statement is false.

Explanation of Solution

Consider the piecewise function $f\left(x\right)=\{\begin{array}{l}2x+1\text{if }x\ne 0\\ 3\text{if }x=0\end{array}$.

From the above example $a=0$ and $f\left(a\right)=3$.

And the value $\underset{x\to a}{\lim }f\left(x\right)$ is calculated as follows.

$\begin{array}{c}\underset{x\to 0}{\lim }f\left(x\right)=2\left(0\right)+1\\ =1\end{array}$

Thus, it is noticeable that the value of $f\left(a\right)=3$ and $\underset{x\to a}{\lim }f\left(x\right)$ is zero at $x=a$.

Therefore, the statement is false.

To determine

Whether the statement “If $\underset{x\to a}{\lim }f\left(x\right)=\infty$ or $\underset{x\to a}{\lim }f\left(x\right)=-\infty$ then $\underset{x\to a}{\lim }f\left(x\right)$ does not exist” is true or false.

Answer to Problem 1RE

The statement is true.

Explanation of Solution

The given limit functions $\underset{x\to a}{\lim }f\left(x\right)=\infty$ or $\underset{x\to a}{\lim }f\left(x\right)=-\infty$ does not approach a finite value

By the definition of the limit, $\underset{x\to a}{\lim }f\left(x\right)$ does not exist.

Hence, the statement is true.

To determine

Whether the statement “If $\underset{x\to a}{\lim }f\left(x\right)$ does not exist, then either $\underset{x\to a}{\lim }f\left(x\right)=\infty$ or $\underset{x\to a}{\lim }f\left(x\right)=-\infty$” is true or false.

Answer to Problem 1RE

The statement is false.

Explanation of Solution

Consider the piecewise function $f\left(x\right)=\{\begin{array}{l}x-2\text{if }x<3\\ {\left(x-5\right)}^2\text{if }x\ge 3\end{array}$ (2)

Take limit in the equation (2) as x approaches 3 from the left.

$\begin{array}{c}\underset{x\to 3^{-}}{\lim }f\left(x\right)=\underset{x\to 3^{-}}{\lim }\left(x-2\right)\\ =3-2\\ =1\end{array}$

Take limit in the equation (2) as x approaches 3 from the right.

$\begin{array}{c}\underset{x\to 3^{+}}{\lim }f\left(x\right)=\underset{x\to 3^{+}}{\lim }{\left(x-5\right)}^2\\ ={\left(3-5\right)}^2\\ ={\left(-2\right)}^2\\ =4\end{array}$

Here $\underset{x\to 3^{-}}{\lim }f\left(x\right)=1$ and $\underset{x\to 3^{+}}{\lim }f\left(x\right)=4$. The values of the right hand side and left hand side limit are not the same. Therefore, $\underset{x\to a}{\lim }f\left(x\right)$ does not exist.

Therefore, the given statement is false.

To determine

Whether the statement “The line $y=2x+1$ is a slant asymptote of the function $f\left(x\right)=\frac{2x^2+x}{x-3}$” is true or false.

Answer to Problem 1RE

The statement is false.

Explanation of Solution

Consider the given function $f\left(x\right)=\frac{2x^2+x}{x-3}$.

The given function can be expressed by using polynomial long division as follows.

$\begin{array}{l}x-3\begin{array}{c}\hfill 2x+7\\ \hfill \overline{)\begin{array}{l}2x^2+x\\ \underset{\_}{2x^2-6x}\end{array}}\end{array}\\ 7x\\ \underset{\_}{7x-21}\\ 21\end{array}$

Therefore, $f\left(x\right)=2x+7+\frac{21}{x-3}$.

Thus, the slant asymptote is $y=2x+7$.

Therefore, the given statement is false.

To determine

Whether the statement “If a function is continuous on the intervals $\left(a,b\right)$ and $\left[b,c\right)$, where $a<b<c$, then the function is also continuous on $\left(a,c\right)$” is true or false.

Answer to Problem 1RE

The statement is false.

Explanation of Solution

Assume that the function continuous on the intervals $\left(0,2\right)$ and $\left[2,5\right)$.

Consider the piecewise function $f\left(x\right)=\{\begin{array}{l}5\text{if }0<x<2\\ 2\text{if }2\le x<5\end{array}$.

From the above example, it is noted that the function is not continuous on $\left(0,5\right)$.

Therefore, the statement is false.

To determine

Whether the statement “If $\underset{x\to a}{\lim }f\left(x\right)$ can be calculated by direct substitution, then f is continuous at $x=a$” is true or false.

Answer to Problem 1RE

The statement is true.

Explanation of Solution

Consider $f\left(x\right)=2x+1$ and $x=2$.

Take limit as x approaches 2.

$\begin{array}{c}\underset{x\to 2}{\lim }f\left(x\right)=\underset{x\to 2}{\lim }\left(2x+1\right)\\ =2\left(2\right)+1\\ =4+1\\ =5\end{array}$

Now substitute $x=2$ in $f\left(x\right)$.

$\begin{array}{c}f\left(2\right)=2\left(2\right)+1\\ =4+1\\ =5\end{array}$

Thus, $\underset{x\to a}{\lim }f\left(x\right)=f\left(a\right)$, where f is continuous at $x=a$.

Hence, the statement is true.