Q. 40 - 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| < $\epsilon$.

$\underset{x\to 1}{\lim }2x^4=2; \mathrm{you} \mathrm{may} \mathrm{assume} \delta \le 1$.

Step-by-Step Solution

Answer

$\delta =min(1,\frac{\epsilon }{30})$

Step1. Given information.

We have been given a limit statement as $\underset{x\to 1}{\lim }2x^4=2; \mathrm{you} \mathrm{may} \mathrm{assume} \delta \le 1$.

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

Step 2. Use algebra.

From the given limit statement, we can identify 

$$\begin{gathered} f\left(x\right)=2x^4 \\ c=1 \\ L=2 \\ \mathrm{For} \epsilon >0 \\ \left|\left(2x^4\right)-2\right|<\epsilon \\ 2\left|x^4-1\right|<\epsilon \\ \left|\left(x^2-1\right)\left(x^2+1\right)\right|<\frac{\epsilon }{2} \\ \left|(x-1)(x+1)\left(x^2+1\right)\right|<\frac{\epsilon }{2} \\ |x-1|\cdot \left|(x+1)\left(x^2+1\right)\right|<\frac{\epsilon }{2} \\ \delta \left|(x+1)\left(x^2+1\right)\right|<\frac{\epsilon }{2} \end{gathered}$$

Step 3. Put x = 2 .

$$\begin{gathered} \delta \cdot \left|(2+1)\left(2^2+1\right)\right|<\frac{\epsilon }{2} \\ \delta \cdot |3\cdot 5|<\frac{\epsilon }{2} \\ \delta \cdot \left|15\right|<\frac{\epsilon }{2} \\ \delta <\frac{\epsilon }{30} \\ \mathrm{For} 0<|x-c|<\delta , \mathrm{we} \mathrm{get} |x-1|<\frac{\epsilon }{30} \\ \mathrm{Therefore}, \delta =min(1,\frac{\epsilon }{30}) \end{gathered}$$