Q1. - Calculus

Question

True/False: 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: If $x\ne c$, then $\left|x-c\right|$ is strictly greater than zero.

(b) True or False: If $\left|x-c\right|$ is strictly greater than zero, then $x\ne c$.

(c) True or False: x is a solution of $0<\left|x-c\right|<\delta$ if and only if $c-\delta <x<c+\delta$.

(d) True or False: If $0<\left|x-c\right|<\delta$, then $x\in \left(c-\delta ,c\right)\cup \left(c,c+\delta \right)$.

(e) True or False: If $\left|f\left(x\right)-L\right|<\in$ then $L-\in <f\left(x\right)<L+\in$.

(f) True or False: If $f\left(x\right)\in \left(L-\in ,L+\in \right)$ then $0<f\left(x\right)<\left|L+\in \right|$.

(g) True or False: The fact that $0<\left|x-3\right|<0.25$ guarantees that $\left|\left(2x-1\right)-5\right|<0.5$ proves that 

(h) True or False: $\underset{x\to 3}{\lim }(2x-1)=5$ means that for all $\delta >0$ there is some $\epsilon >0$ such that if $0<\left|x-c\right|<\delta$, then $\left|(2x-1)-5\right|<\epsilon$.

Step-by-Step Solution

Answer

Part (a) True

Part (b) True

Part (c) False

Part (d) True

Part (e) True

Part (f) False 

Part (g) True

Part (h) True 

Part (a) Step 1. Explanation.

Consider the given function,

Since the mod function gives always a positive value. So, for any value of c, the function is strictly greater than zero.

Part (b) Step 1. Explanation.

Consider the given information.

If x is equal to c, then the value of $\left|x-c\right|$ will be zero. So the function is strictly greater than zero if and only if $x\ne c$. Thus, the given statement is true.

Part (c) Step 1. Explanation.

Consider the given information.

Let's take the given value.

$\begin{array}{l}\left|x-c\right|<\delta \\ -\delta <x-c<\delta \\ c-\delta <x<\delta +c\end{array}$

It's an expansion of the mod inequality. But while using zero, this implies that all the values should be greater than zero. So, the given statement is not true, while x is the solution.

Part (d) Step 1. Explanation.

Consider the given information:

$\begin{array}{l}0<\left|x-c\right|<\delta \\ -\delta <x-c<\delta \\ c-\delta <x<\delta +c\end{array}$

This implies that x is not equal to zero. Thus, the value of x will lie between the interval $x\in \left(c-\delta ,c\right)\cup \left(c,c+\delta \right)$.

Thus, the given statement is true.

Part (e) Step 1. Explanation.`

Consider the given information.

The given statement is true, As this is the basic rule to open the modulus function.

$\begin{array}{rcl}\left|f\left(x\right)-c\right|& <& \delta \\ -\delta & <& f\left(x\right)-c<\delta \\ c-\delta & <& f\left(x\right)<\delta +c\end{array}$

Part (f) Step 1. Given information.

Consider the given information.

The given statement is false as,

$\begin{array}{rcl}0& <& f\left(x\right)<\left|L+\in \right|\\ -L-& \in & <f\left(x\right)<L+\in \end{array}$

While putting a value in both given functions, we are getting different values. So, the given statement is not true.

Part (g) Step 1. Given information.

Consider the given information,

Simplify both the given inequality.

$\begin{array}{l}0<\left|x-3\right|<0.25\\ -0.25<x-3<0.25\\ 3-0.25<x<0.25+3\\ 2.75<x<3.25\end{array}$

And,

$\begin{array}{l}0<\left|\left(2x-1\right)-5\right|<0.5\\ -0.5<\left(2x-1\right)-5<0.5\\ -0.5<2x-6<0.5\\ 5.5<2x<6.5\\ 2.75<2x<3.25\end{array}$

Thus, both the value of inequalities is the same. Thus, the statement is true.

Part (h) Step 1. Given information.

Consider the given values:

$\underset{x\to c}{\lim }f\left(x\right)=L\iff \forall \in >0,\exists \delta >0\mathrm{s}.\mathrm{t}.0<|x-c|<\delta \to \left|f\right(x)-L|<\in$

And,

$\begin{array}{l}c=3\\ L=5\\ \in =0.5\\ f\left(x\right)=2x-1\end{array}$

From the definitions of limits, the given statement is true.