Q. 18 - Calculus

Question

Show that the limit as $x\to 2 \mathrm{of} \mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}-1.1}$ is not equal to 1, by finding an  $\epsilon >0$ for which there is no corresponding $\delta >0$ satisfying the formal definition of limit.

Step-by-Step Solution

Answer

The $\delta <0$, the expression is false.

Step 1. Given information.

The given function is $\underset{x\to 2}{\lim }\sqrt{x-1.1}\ne 1$.

Step 2. Calculation.

From the given expression, we have, $c=2, L=1$.

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

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

$x\in \left(2-\delta ,2\right)\cup \left(2,2+\delta \right) \mathrm{then} \sqrt{\mathrm{x}-1.1}\in \left(1-\mathrm{\epsilon },1+\mathrm{\epsilon }\right)$.

So the largest value of $\delta$ is:

$$\begin{gathered} \delta =\left(1^2-1.1\right)-2 \\ \delta =1-1.1-2 \\ \delta =1-3.1 \\ \delta =-2.1 \end{gathered}$$

Since $\delta <0$, the expression is false.