Question

evaluate the limit
if the limit does not exist enter dne.
limit =
lim_{y
ightarrow4}\frac{6(y^{2}-1)}{3y^{2}(y - 1)^{3}}

Explanation:

Step1: Factor the numerator

Use the difference - of - squares formula \(a^{2}-b^{2}=(a + b)(a - b)\). Here, \(y^{2}-1=(y + 1)(y - 1)\), so the function becomes \(\lim_{y
ightarrow4}\frac{6(y + 1)(y - 1)}{3y^{2}(y - 1)^{3}}\).

Step2: Simplify the function

Cancel out the common factor \((y - 1)\) in the numerator and denominator. We get \(\lim_{y
ightarrow4}\frac{6(y + 1)}{3y^{2}(y - 1)^{2}}\).

Step3: Substitute \(y = 4\)

Substitute \(y = 4\) into the simplified function \(\frac{6(y + 1)}{3y^{2}(y - 1)^{2}}\).
\[

$$\begin{align*} \frac{6(4 + 1)}{3\times4^{2}\times(4 - 1)^{2}}&=\frac{6\times5}{3\times16\times9}\\ &=\frac{30}{432}\\ &=\frac{5}{72} \end{align*}$$

\]

Answer:

\(\frac{5}{72}\)