Chapter 11.3, Problem 17E

Solution

a)

To explain: Why you cannot use substitution to find the limit $\underset{x\to 0}{\lim }\sqrt{x-3}$ .

When one substitutes 0 for $x$ then the radicand becomes negative, so one cannot use substitution to find the limit $\underset{x\to 0}{\lim }\sqrt{x-3}$ .

Given information: The given limit is $\underset{x\to 0}{\lim }\sqrt{x-3}$ .

Calculation:

Substitute $0$ for $x$ in the given limit then the denominator becomes zero as shown below:

  $\begin{array}{c}\sqrt{x-3}=\sqrt{0-3}\\ =\sqrt{-3}\end{array}$

Thus, the radicand becomes negative so the square root is not defined, so one cannot use substitution to find the given limit.

b)

To find: The limit $\underset{x\to 0}{\lim }\sqrt{x-3}$ algebraically if it exists.

The limit $\underset{x\to 0}{\lim }\sqrt{x-3}$ does not exists and thus it cannot be found.

Given information: The given limit is $\underset{x\to 0}{\lim }\sqrt{x-3}$ .

Calculation:

The domain of $y=\sqrt{x-3}$ is $\left[3,\infty \right)$ and the graph lies on this interval. Thus the limit does not exists and cannot be found.