Question

$\lim\limits_{y \to 3} \frac{7(y^2 - 1)}{8y^2(y - 1)^3}$

Explanation:

Step1: Substitute \( y = 3 \) into the function

We substitute \( y = 3 \) into the numerator \( 7(y^2 - 1) \) and the denominator \( 8y^2(y - 1)^3 \) respectively.
For the numerator: \( 7(3^2 - 1)=7\times(9 - 1)=7\times8 = 56 \)
For the denominator: \( 8\times3^2\times(3 - 1)^3=8\times9\times2^3=8\times9\times8 = 576 \)

Step2: Calculate the limit

Since the function is continuous at \( y = 3 \) (the denominator is not zero when \( y = 3 \)), the limit is the value of the function at \( y = 3 \), so \( \lim\limits_{y
ightarrow3}\frac{7(y^2 - 1)}{8y^2(y - 1)^3}=\frac{56}{576} \)
Simplify the fraction: \( \frac{56\div8}{576\div8}=\frac{7}{72} \)

Answer:

\( \frac{7}{72} \)