Q. 38 - 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| < $\epsilon$.

$\underset{x\to 3}{\lim }\frac{1}{x}=\frac{1}{3}; \mathrm{you} \mathrm{may} \mathrm{assume} \delta \le 1$

Step-by-Step Solution

Answer

$\delta =min(1,3\epsilon )$

Step1. Given information.

We have been given a limit statement as $\underset{x\to 3}{\lim }\frac{1}{x}=\frac{1}{3}; \mathrm{you} \mathrm{may} \mathrm{assume} \delta \le 1$.

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

Step 2. Use algebra.

From the given limit statement, we can identify 

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