Question

find the following limit.
lim_{x
ightarrow - 0.5^{+}}sqrt{\frac{x + 5}{x + 2}}
lim_{x
ightarrow - 0.5^{+}}sqrt{\frac{x + 5}{x + 2}}=square

Explanation:

Step1: Substitute the value of x

We directly substitute \(x = - 0.5\) into the function \(\sqrt{\frac{x + 5}{x+2}}\) since the function \(\sqrt{\frac{x + 5}{x + 2}}\) is continuous at \(x=-0.5\).
When \(x=-0.5\), we have \(\frac{x + 5}{x+2}=\frac{-0.5 + 5}{-0.5+2}\).

Step2: Calculate the fraction inside the square - root

First, calculate the numerator: \(-0.5 + 5=4.5\).
Then, calculate the denominator: \(-0.5 + 2 = 1.5\).
So, \(\frac{-0.5 + 5}{-0.5+2}=\frac{4.5}{1.5}=3\).

Step3: Calculate the square - root

We have \(\sqrt{\frac{-0.5 + 5}{-0.5+2}}=\sqrt{3}\).

Answer:

\(\sqrt{3}\)