Q. 32 - Calculus

Question

For each limit statement , use algebra to find δ > 0 in terms of $\epsilon$ > 0 so that if 0 < |x − c| < δ, then | f(x) − L| < $\epsilon$.

$\underset{x\to 1}{\lim }(3x+8)=11$

Step-by-Step Solution

Answer

$\delta =\frac{\epsilon }{3}$

Step1. Given information.

We have been given a limit statement as $\underset{x\to 1}{\lim }(3x+8)=11$.

We have to find $\delta \mathrm{in} \mathrm{terms} \mathrm{of} \epsilon$.

Step 2. Use algebra

From the given limit statement, we can identify 

$$\begin{gathered} f\left(x\right)=3x+8 \\ c=1 \\ L=11 \\ \mathrm{For} \epsilon >0 \\ \left|\right(3x+8)-11|<\epsilon \\ |3x+8-11|<\epsilon \\ |3x-3|<\epsilon \\ 3|x-1|<\epsilon \\ |x-1|<\frac{\epsilon }{3} \\ \mathrm{For} 0<|x-c|<\delta , \mathrm{we} \mathrm{get} |x-1|<\frac{\epsilon }{3} \\ \mathrm{Therefore}, \delta =\frac{\epsilon }{3} \end{gathered}$$