Question

find the following limit or state that it does not exist. (limlimits_{x \to 49} \frac{sqrt{x} - 7}{x - 49}) select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. (limlimits_{x \to 49} \frac{sqrt{x} - 7}{x - 49} = ) (type an exact answer) b. the limit does not exist.

Explanation:

Step1: Factor the denominator

Notice that \(x - 49\) is a difference of squares, so \(x - 49=(\sqrt{x})^2 - 7^2 = (\sqrt{x}-7)(\sqrt{x}+7)\).
The limit becomes \(\lim_{x
ightarrow49}\frac{\sqrt{x}-7}{(\sqrt{x}-7)(\sqrt{x}+7)}\).

Step2: Cancel the common factor

As \(x
ightarrow49\), \(\sqrt{x}
eq7\) (we are taking the limit, not evaluating at \(x = 49\) directly, but as \(x\) approaches \(49\), \(\sqrt{x}\) approaches \(7\) but is not equal to \(7\) in the neighborhood around \(49\) except at \(49\) itself, so we can cancel \(\sqrt{x}-7\) from the numerator and the denominator).
After canceling, we get \(\lim_{x
ightarrow49}\frac{1}{\sqrt{x}+7}\).

Step3: Substitute \(x = 49\)

Now we can substitute \(x = 49\) into the function \(\frac{1}{\sqrt{x}+7}\) because the function is now defined at \(x = 49\) (the denominator \(\sqrt{49}+7=7 + 7=14
eq0\)).
So \(\frac{1}{\sqrt{49}+7}=\frac{1}{14}\).

Answer:

\(\frac{1}{14}\)