QuestionDetermining Limits Determine the following limits: a) $\quad \lim _{x \rightarrow 2}\left(x^{2}-4\right)$ j) $\quad \lim _{x \rightarrow 2} \frac{x-2}{x^{2}-3 x+2}$ s) $\lim _{x \rightarrow-2} \frac{x^{3}+3 x^{2}+2 x}{x^{2}-x-6}$ b) $\lim _{x \rightarrow 0} \frac{x^{3}-4 x}{2 x^{2}+3 x}$ k) $\lim _{x \rightarrow 0} \frac{3 x+2 x^{-1}}{x+4 x^{-1}}$ t) $\lim _{x \rightarrow 1} \frac{x^{2}-2 x+1}{x^{3}-x}$

Studdy Solution
Substitute $x = 1$:
$$\frac{1-1}{1(1+1)} = \frac{0}{2} = 0$$
So,
$$\lim_{x \rightarrow 1} \frac{x^2 - 2x + 1}{x^3 - x} = 0$$
The solutions to the limits are:
a) $\lim_{x \rightarrow 2} (x^2 - 4) = 0$
j) $\lim_{x \rightarrow 2} \frac{x-2}{x^2 - 3x + 2} = 1$
s) $\lim_{x \rightarrow -2} \frac{x^3 + 3x^2 + 2x}{x^2 - x - 6} = \frac{2}{5}$
b) $\lim_{x \rightarrow 0} \frac{x^3 - 4x}{2x^2 + 3x} = \frac{-4}{3}$
k) $\lim_{x \rightarrow 0} \frac{3x + 2x^{-1}}{x + 4x^{-1}} = \frac{1}{2}$
t) $\lim_{x \rightarrow 1} \frac{x^2 - 2x + 1}{x^3 - x} = 0$