Q. 1C - Calculus

Question

Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a counterexample.

a. True or False: The limit of a difference of functions as \(x\rightarrow c\) is equal to the difference of the limits of those functions as \(x\rightarrow c\) , provided that all limits involved exist.

b. True or False: If \(f(x)\) is within \(0.25\) unit of \(7\) and \(g(x)\) is within \(0.25\) unit of \(2\), then \(f(x) + g(x)\) is within \(0.5\) unit of \(9\).

c.  True or False: If \(f(x)\) is within \(0.25\) unit of \(7\) and \(g(x)\) is within \(0.25\) unit of \(2\), then \(f(x)g(x)\) is within \(0.5\) unit of \(9\).

d. True or False: Every algebraic function \(f\) is continuous at every real number \(x = c\)

e. True or False: Every power function \(f(x) = Ax^{k}\) is continuous at the point \(x = 2\). 

f. True or False: The function \(f(x) = sec x\) is continuous at \(x=\frac{\pi }{2}\).

g. True or False: The value of \(\frac{(x-c)f(x)}{(x-c)g(x)}\) at \(x = c\) is equal to the limit of \(\frac{f(x)}{g(x)}\) at \(x=x\).

h. True or False: The limit of \(\frac{(x-c)f(x)}{(x-c)g(x)}\) as \(x\rightarrow\) is equal to the limit of \(\frac{f(x)}{g(x)}\) as \(x\rightarrow c\).

Step-by-Step Solution

Answer

Part a. True.

Part b. True.

Part c. False.

Part d. False.

Part e. True.

Part f. False.

Part g. False.

Part h. True.

Part a. Step 1. Explanation

It is known that,

\(\lim_{x\rightarrow c}{f(x)-g(x)}=\lim_{x\rightarrow c}f(x)-\lim_{x\rightarrow c}g(x)\): provided that all limit exist.

Hence,

“The limit of a difference of functions as \(x\rightarrow c\) is equal to the difference of the limits of those function as\(x\rightarrow c\) , provided that all limit exist.”

Given statement is true.

Part b. Step 1. Given information

Given,

\(6.75<\lim_{x\rightarrow c}f(x)<7.25\)

\(1.75<\lim_{x\rightarrow c}g(x)<2.25\)

Part b. Step 2. Explanation

From above inequalities,

\(8.50<\lim_{x\rightarrow c}f(x)+\lim_{x\rightarrow c} g(x)<9.50\)

This implies that if \(f(x)\) is within \(0.25\) unit of \(7\) & \(g(x)\) is within \(0.25\) unit of \(2\), then \(f(x)+g(x)\) is within \(0.5\) unit of \(9\).

Given statement is true.

Part c. Step 1. Explanation

From above inequalities,

\(\lim_{x\rightarrow c}f(x).\lim_{x\rightarrow c}g(x)<(7.25).(2.25)\).

Hence, the given statement is false.

Part d. Step 1. Explanation

Let’s consider a function,

\(f(x)=\frac{1}{x-1}\)

Above function is an algebraic function discontinuous at \(x=1\)

This does not implies that every algebraic function \(f\) is continuous at every real number \(x=c\)

 

Given statement is false.

Part e. Step 1. Explanation

Let’s consider the graph of \(x,x^{2},x^{3},x^{-1},x^{-2},x^{-3}\)

It can be clearly understood that there would be not point of discontinuity at \(x=2\) for \(f(x)=Ax^{k}\).

Part f. Step 1. Explanation

It is known that \(f(x)\) is undefined at \(x=\frac{\pi }{2}\). Therefore function is undefined at \(x=\frac{\pi }{2}\).

Given statement is false.

Part g. Step 1. Explanation

\(\frac{(x-c)f(x)}{(x-c)g(x)}\)

Term \((x-c)\) will get cancelled from numerator and denominator and its value will be \(\frac{f(x)}{g(x)}\) at \(x=c\), provided that value of \(f(x)\) & \(g(x)\) exist at \(x=c\). Hence there is no meaning of limit of \(\frac{f(x)}{g(x)}\).

Given statement is false.

Part h. Step 1. Explanation

Term \((x-c)\) will get cancelled from numerator and denominator and its value will be limit of \(\frac{f(x)}{g(x)}\) as \(x\rightarrow c\) provided that value of \(f(x)\) & \(g(x)\) exists as \(x\rightarrow c\).

Hence, given statement is true.