Question

use lhôpitals rule to find the limit.

$$\lim_{y\to 0}\frac{\sqrt{5y + 16}-4}{y}$$

$$\lim_{y\to 0}\frac{\sqrt{5y + 16}-4}{y}=\square$$ (type an integer or a simplified fraction.)

Explanation:

Step1: Check the form of the limit

When \(y = 0\), \(\frac{\sqrt{5y + 16}-4}{y}=\frac{\sqrt{16}-4}{0}=\frac{0}{0}\), which is an indeterminate form. So, we can apply L'Hôpital's Rule.

Step2: Differentiate the numerator and the denominator

The derivative of the numerator \(u=\sqrt{5y + 16}-4=(5y + 16)^{\frac{1}{2}}-4\) using the chain rule \((u^n)^\prime=nu^{n - 1}u^\prime\). Let \(t = 5y+16\), then \(u=t^{\frac{1}{2}}-4\). \(u^\prime=\frac{1}{2}(5y + 16)^{-\frac{1}{2}}\times5=\frac{5}{2\sqrt{5y + 16}}\).
The derivative of the denominator \(v = y\) is \(v^\prime=1\).
By L'Hôpital's Rule, \(\lim_{y
ightarrow0}\frac{\sqrt{5y + 16}-4}{y}=\lim_{y
ightarrow0}\frac{\frac{5}{2\sqrt{5y + 16}}}{1}\).

Step3: Evaluate the new limit

Substitute \(y = 0\) into \(\frac{5}{2\sqrt{5y + 16}}\). We get \(\frac{5}{2\sqrt{16}}=\frac{5}{2\times4}\).

Answer:

\(\frac{5}{8}\)