Q. 45 - Calculus

Question

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

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

Step-by-Step Solution

Answer

The largest value of $\delta =3$.

Step 1. Given Information.

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

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

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

Now, the largest value of delta is given by 

$$\begin{gathered} \sqrt{x+1}=L+\epsilon \\ \sqrt{x+1}=2+1 \\ x+1=3^2 \\ x+1=9 \\ x=8 \end{gathered}$$

Thus,

$$\begin{gathered} \delta =8-5 \\ \delta =3 \end{gathered}$$

Step 3. Conclusion.

The largest value of $\delta =3$.