Q. 22 - Calculus

Question

Suppose $f\left(x\right)=mx+b$is a linear function with $m\ne 0$, and let c be any real member.

Part (a): Show that for all $\in >0$, if $0<\left|x-c\right|<\frac{\in }{\left|m\right|}$, then $\left|f\left(x\right)-f\left(c\right)\right|<\in$.

Part (b): What does the implication in part (a) have to do with limits?

Part (c): Illustrate the implication in part (a) with a labeled graph. Explain in terms of slopes why it makes sense that given $\in >0$, the corresponding $\delta >0$ is $\delta =\frac{\in }{\left|m\right|}$.

Step-by-Step Solution

Answer

Part (a): To prove $\left|f\left(x\right)-f\left(c\right)\right|<\in$, put $x=c$ and multiply $\left|m\right|$ on every side and solve the inequation.

Part (b): Using limits, we get $\underset{x\to c}{\lim }x=c$.

Part (c): The required graph is given below,

Part (a) Step 1. Given information.

Consider the given question,

$$\begin{gathered} f\left(x\right)=mx+b \\ \in >0 \end{gathered}$$

If $0<\left|x-c\right|<\frac{\in }{\left|m\right|}$ then $\left|f\left(x\right)-f\left(c\right)\right|<\in$.

Part (a) Step 2. Substitute c in x in the given function.

Substitute $x=c$ in the given function,

$f\left(c\right)=mc+c$

For all $\in >0$,

$0<\left|x-c\right|<\frac{\in }{\left|m\right|}$

Multiply $\left|m\right|$ on every side,

$$\begin{gathered} 0·\left|m\right|<\left|x-c\right|·\left|m\right|<\frac{\in }{\left|m\right|}·\left|m\right| \\ 0<\left|mx-mc\right|<\in \\ 0<\left|\left(mx+c\right)-\left(mc+c\right)\right|<\in \\ 0<\left|f\left(x\right)-f\left(c\right)\right|<\in \end{gathered}$$

Therefore, $\left|f\left(x\right)-f\left(c\right)\right|<\in$.

Hence, proved.

Part (b) Step 1. To defined the given implication with limits.

From the definition of limit,

If $0<\left|x-c\right|<\frac{\in }{\left|m\right|}$,

Then, it can be written,

$\underset{x\to c}{\lim }x=c$

Part (c) Step 1. Sketch the graph.

On sketching the graph, we get,