Q. 16 - Calculus

Question

It is false that $\underset{x\to 1}{\lim }\left(x+1.01\right)=2$. Express this fact in a mathematical sentence involving $\delta$ and $\epsilon$, to show how the formal definition of limit fails in this case.

Step-by-Step Solution

Answer

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

Step 1. Given information.

The given function is $\underset{x\to 1}{\lim }\left(x+1.01\right)=2$.

Step 2. Calculation.

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

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

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

$x\in \left(1-\delta ,1\right)\cup \left(1,1+\delta \right) \mathrm{then} \mathrm{x}+1.01\in \left(2-\mathrm{\epsilon },2+\mathrm{\epsilon }\right)$.

So, the largest value of $\delta$ is given by:

$$\begin{gathered} \delta =\left(2-1.01\right)-1 \\ \delta =2-1.01-1 \\ \delta =2-2.01 \\ \delta =-0.01 \end{gathered}$$

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

Step 3. Conclusion.

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