Question

1(b) $lim_{y
ightarrow - 4}\frac{y^{3}-64}{y^{2}-7y + 12}$

Explanation:

Step1: Factor the numerator and denominator

The numerator $y^{3}-64$ is a difference - of - cubes and can be factored as $(y - 4)(y^{2}+4y + 16)$ using the formula $a^{3}-b^{3}=(a - b)(a^{2}+ab + b^{2})$ where $a = y$ and $b = 4$. The denominator $y^{2}-7y + 12$ can be factored as $(y - 3)(y - 4)$ using the formula $x^{2}+(a + b)x+ab=(x + a)(x + b)$ where $a=-3$ and $b=-4$. So the limit becomes $\lim_{y
ightarrow4}\frac{(y - 4)(y^{2}+4y + 16)}{(y - 3)(y - 4)}$.

Step2: Cancel out the common factor

Since $y
ightarrow4$ but $y
eq4$, we can cancel out the common factor $(y - 4)$ in the numerator and denominator. The limit simplifies to $\lim_{y
ightarrow4}\frac{y^{2}+4y + 16}{y - 3}$.

Step3: Substitute the value of $y$

Substitute $y = 4$ into the simplified function $\frac{y^{2}+4y + 16}{y - 3}$. We get $\frac{4^{2}+4\times4 + 16}{4 - 3}=\frac{16+16 + 16}{1}=48$.

Answer:

$48$