Question

1. $lim_{x

ightarrow5}sqrt{3x - 11}$ 3. $lim_{x
ightarrow0}(-5)$

Explanation:

Step1: Substitute the value of x

We use the direct - substitution property of limits. If \(y = f(x)\) is continuous at \(x = a\), then \(\lim_{x
ightarrow a}f(x)=f(a)\). The function \(y = \sqrt{3x - 11}\) is continuous for \(3x-11\geq0\). When \(x = 5\), we substitute \(x = 5\) into \(\sqrt{3x - 11}\).
\(\sqrt{3\times5 - 11}\)

Step2: Simplify the expression

First, calculate the value inside the square - root: \(3\times5-11=15 - 11 = 4\). Then, \(\sqrt{4}=2\).
For \(\lim_{x
ightarrow0}(- 5)\), since the function \(y=-5\) is a constant function, and for a constant function \(y = c\), \(\lim_{x
ightarrow a}c=c\) for any real number \(a\). So \(\lim_{x
ightarrow0}(-5)=-5\).

Answer:

\(\lim_{x
ightarrow5}\sqrt{3x - 11}=2\), \(\lim_{x
ightarrow0}(-5)=-5\)