Q. 39 - 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| < $\epsilon$.

$\underset{x\to 3}{\lim }(x^2-2x-3)=0; \mathrm{you} \mathrm{may} \mathrm{assume} \delta \le 1$

Step-by-Step Solution

Answer

$\delta =min(1,\frac{\epsilon }{5})$

Step1. Given information.

We have been given a limit statement as $\underset{x\to 3}{\lim }(x^2-2x-3)=0; \mathrm{you} \mathrm{may} \mathrm{assume} \delta \le 1$.

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

Step 2. Use algebra.

From the given limit statement, we can identify 

$$\begin{gathered} f\left(x\right)=x^2-2x-3 \\ c=3 \\ L=0 \\ \mathrm{For} \epsilon >0 \\ \left|\left(x^2-2x-3\right)-0\right|<\epsilon \\ \left|x^2-3x+x-3\right|<\epsilon \\ \left|x\right(x-3)+1(x-3\left)\right|<\epsilon \\ \left|\right(x-3\left)\right(x+1\left)\right|<\epsilon \\ |x-3|\cdot |x+1|<\epsilon \\ \delta |x+1|<\epsilon \end{gathered}$$

Step 3. Put x = 4

$$\begin{gathered} \delta \cdot |4+1|<\epsilon \\ \delta \cdot \left|5\right|<\epsilon \\ \delta <\frac{\epsilon }{5} \\ \mathrm{For} 0<|x-c|<\delta , \mathrm{we} \mathrm{get} |x-3|<\frac{\epsilon }{5} \\ \mathrm{Therefore}, \delta =min(1,\frac{\epsilon }{5}) \end{gathered}$$