use lhopitals rule to find the limit. lim y→0 (√(3y + 16) - 4)/y lim y→0 (√(3y + 16) - 4)/y = (type an…

Explanation:

Step1: Check indeterminate form

When (y = 0), (\frac{\sqrt{3y + 16}-4}{y}=\frac{\sqrt{16}-4}{0}=\frac{4 - 4}{0}=\frac{0}{0}), so L'Hopital's Rule can be applied.

Step2: Differentiate numerator and denominator

The derivative of the numerator (u=\sqrt{3y + 16}-4=(3y + 16)^{\frac{1}{2}}-4), using the chain - rule (u^\prime=\frac{3}{2\sqrt{3y + 16}}). The derivative of the denominator (v = y) is (v^\prime=1).

Step3: Find the new limit

(\lim_{y\rightarrow0}\frac{\frac{3}{2\sqrt{3y + 16}}}{1}=\lim_{y\rightarrow0}\frac{3}{2\sqrt{3y + 16}}).

Step4: Evaluate the limit

Substitute (y = 0) into (\frac{3}{2\sqrt{3y + 16}}), we get (\frac{3}{2\sqrt{16}}=\frac{3}{2\times4}=\frac{3}{8}).

Answer:

(\frac{3}{8})