Q. 80 - Calculus

Question

Prove that for all x within 0.01 of the value x = 1, the quantity ${\left(x-1\right)}^2$ is within the interval (0, 0.0001). What does this have to do with $\underset{x\to 1}{\lim }{\left(x-1\right)}^2?$

Step-by-Step Solution

Answer

The given statement is proved. The value of the limit $\underset{x\to 1}{\lim }{\left(x-1\right)}^2=0.$

Step 1. Given Information.

The given quantity is ${\left(x-1\right)}^2.$

Step 2. Prove.

Let the function is $f\left(x\right)={\left(x-1\right)}^2.$

Take the limit of the above function as $x\to 0.01:$

$$\begin{gathered} \underset{x\to 0.01}{\lim }f\left(x\right)=\underset{x\to 0.01}{\lim }{\left(x-1\right)}^2 \\ \underset{x\to 0.01}{\lim }f\left(x\right)={\left(0.01-1\right)}^2 \\ \underset{x\to 0.01}{\lim }f\left(x\right)={\left(-0.99\right)}^2 \\ \underset{x\to 0.01}{\lim }f\left(x\right)=0.9801 \end{gathered}$$

Now, take the limit of the above function as $x\to 1:$

$$\begin{gathered} \underset{x\to 1}{\lim }f\left(x\right)=\underset{x\to 1}{\lim }{\left(x-1\right)}^2 \\ \underset{x\to 1}{\lim }f\left(x\right)={\left(1-1\right)}^2 \\ \underset{x\to 1}{\lim }f\left(x\right)=0 \end{gathered}$$

Therefore, the quantity ${\left(x-1\right)}^2$ is within the interval $\left(0,0.0001\right).$

Step 3. Finding the limit.

Let's find the limit.

$$\begin{gathered} \underset{x\to 1}{\lim }{\left(x-1\right)}^2 \\ ={\left(1-1\right)}^2 \\ =0 \end{gathered}$$