Q. 46 - Calculus

Question

For each limit $\underset{\mathrm{x}\to \mathrm{c}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\mathrm{L}$in Exercises 43–54, use graphs and algebra to approximate the largest value of $\delta$such that if $x\in (c-\delta ,c)\cup (c,c+\delta ) \mathrm{then} \mathrm{f}\left(\mathrm{x}\right)\in (\mathrm{L}-\epsilon ,\mathrm{L}+\epsilon )\mathit{ }a$.

$\underset{\mathrm{x}\to 5}{\lim }\sqrt{\mathrm{x}-1}=2, \mathrm{\epsilon }=0.2$

Step-by-Step Solution

Answer

The largest value of $\delta =1.16$.

Step 1. Given Information.

The given expression is $\underset{\mathrm{x}\to 5}{\lim }\sqrt{\mathrm{x}-1}=2, \mathrm{\epsilon }=0.2$.

Step 2. Explanation.

From the given expression, we have, $c=5, L=2$.

The limit expression can be written as a formal statement as below,

For all epsilon positive, there exists a delta positive such that if

$x\in (5-\delta ,5)\cup (5,5+\delta ) \mathrm{then} \sqrt{x-1}\in (2-\epsilon ,2+\epsilon )\mathit{ }$

Now, the largest value of delta is given by 

$$\begin{gathered} \sqrt{x+1}=L+\epsilon \\ \sqrt{x+1}=2+0.2 \\ x+1={\left(2.2\right)}^2 \\ x+1=4.84 \\ x=4.84-1 \\ x=3.84 \end{gathered}$$

Thus,

$$\begin{gathered} \delta =5-3.84 \\ \delta =1.16 \end{gathered}$$

Step 3. Conclusion.

The largest value of$\delta =1.16$.