use lhôpitals rule to find the limit.\n\n$$\\lim_{y \\to 0} \\frac{\\sqrt{5y + 16} - 4}{y}$$\n\n$$\\lim_{y…

use lhôpitals rule to find the limit.\n\n$$\\lim_{y \\to 0} \\frac{\\sqrt{5y + 16} - 4}{y}$$\n\n$$\\lim_{y \\to 0} \\frac{\\sqrt{5y + 16} - 4}{y} = \\square$$ (type an integer or a simplified fraction.)

use lhôpitals rule to find the limit.\n\n$$\\lim_{y \\to 0} \\frac{\\sqrt{5y + 16} - 4}{y}$$\n\n$$\\lim_{y \\to 0} \\frac{\\sqrt{5y + 16} - 4}{y} = \\square$$ (type an integer or a simplified fraction.)

Answer

Explanation:

Step1: Check the form of the limit

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

Step2: Differentiate the numerator and 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).

Step3: Apply L'Hôpital's Rule

By L'Hôpital's Rule (\lim_{y\rightarrow0}\frac{\sqrt{5y + 16}-4}{y}=\lim_{y\rightarrow0}\frac{\frac{5}{2\sqrt{5y + 16}}}{1})

Step4: Evaluate the 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})