Question

use lhopitals rule to find the limit.
lim (sqrt(5y + 4)-2)/y (y->0)
lim (sqrt(5y + 4)-2)/y = (type an integer or a simplified fraction.) (y->0)

Explanation:

Step1: Check the form of the limit

When \(y = 0\), the numerator \(\sqrt{5y + 4}-2=\sqrt{4}-2=0\) and the denominator \(y = 0\). So, it is in the \(\frac{0}{0}\) form, and we can apply L'Hôpital's Rule.
L'Hôpital's Rule states that if \(\lim_{y
ightarrow a}\frac{f(y)}{g(y)}\) is in the \(\frac{0}{0}\) or \(\frac{\pm\infty}{\pm\infty}\) form, then \(\lim_{y
ightarrow a}\frac{f(y)}{g(y)}=\lim_{y
ightarrow a}\frac{f^{\prime}(y)}{g^{\prime}(y)}\)

Let \(f(y)=\sqrt{5y + 4}-2=(5y + 4)^{\frac{1}{2}}-2\) and \(g(y)=y\)

First, find the derivative of \(f(y)\) using the chain rule. The derivative of \((5y + 4)^{\frac{1}{2}}\) is \(\frac{1}{2}(5y + 4)^{-\frac{1}{2}}\times5=\frac{5}{2\sqrt{5y+4}}\), and the derivative of \(- 2\) is \(0\). So, \(f^{\prime}(y)=\frac{5}{2\sqrt{5y + 4}}\)

The derivative of \(g(y)=y\) is \(g^{\prime}(y) = 1\)

Step2: Apply L'Hôpital's Rule

\(\lim_{y
ightarrow0}\frac{\sqrt{5y + 4}-2}{y}=\lim_{y
ightarrow0}\frac{\frac{5}{2\sqrt{5y+4}}}{1}\)

Step3: Evaluate the new limit

Substitute \(y = 0\) into \(\frac{5}{2\sqrt{5y+4}}\)

When \(y = 0\), we have \(\frac{5}{2\sqrt{5\times0 + 4}}=\frac{5}{2\times2}=\frac{5}{4}\)

Answer:

\(\frac{5}{4}\)