Question

1. a. $lim_{x

ightarrow0}\frac{x - 3}{x^{4}-9x^{2}}$ b. $lim_{x
ightarrow3}\frac{x - 3}{x^{4}-9x^{2}}$ c. $lim_{x
ightarrow - 3}\frac{x - 3}{x^{4}-9x^{2}}$

Explanation:

Step1: Factor the denominator

First, factor $x^{4}-9x^{2}=x^{2}(x^{2} - 9)=x^{2}(x - 3)(x + 3)$. So the function becomes $\frac{x - 3}{x^{2}(x - 3)(x + 3)}=\frac{1}{x^{2}(x + 3)}$ for $x
eq3$.

Step2: Calculate limit as $x

ightarrow0$
Substitute $x = 0$ into $\frac{1}{x^{2}(x + 3)}$. We have $\lim_{x
ightarrow0}\frac{x - 3}{x^{4}-9x^{2}}=\lim_{x
ightarrow0}\frac{1}{x^{2}(x + 3)}=\infty$ since the denominator approaches $0$ and the numerator is non - zero.

Step3: Calculate limit as $x

ightarrow3$
Substitute $x = 3$ into $\frac{1}{x^{2}(x + 3)}$. $\lim_{x
ightarrow3}\frac{x - 3}{x^{4}-9x^{2}}=\lim_{x
ightarrow3}\frac{1}{x^{2}(x + 3)}=\frac{1}{3^{2}(3 + 3)}=\frac{1}{9\times6}=\frac{1}{54}$.

Step4: Calculate limit as $x

ightarrow - 3$
Substitute $x=-3$ into $\frac{1}{x^{2}(x + 3)}$. The denominator $x^{2}(x + 3)$ approaches $0$ and the numerator is non - zero, so $\lim_{x
ightarrow - 3}\frac{x - 3}{x^{4}-9x^{2}}=\infty$.

Answer:

a. $\infty$
b. $\frac{1}{54}$
c. $\infty$