Solve limit of rational function

Question

I have function: $$\lim\limits_{x\rightarrow -2}\frac{x^3+3x^2+2x}{x^2-x-6}$$ after calculations I got: $$\frac{-8+12-4}{4+2-6} = \frac{0}{0}$$ What must I do from here? I have tried to divide by (x-1), but I guess this is wrong. Can you post all steps to answer with explanation. Thank you

Answer 1

We have $\require{cancel}x^3 + 3x^2 + 2x = x(x^2 + 3x + 2) = x(x+1)(x+2)$. Also, $x^2-x-6 = (x-3)(x+2)$, so

$$\begin{align} \lim\limits_{x\rightarrow -2}\frac{x^3+3x^2+2x}{x^2-x-6} & = \lim_{x \to -2} \frac{x(x+1)(x+2)}{(x-3)(x+2)} \\ & = \lim_{x \to -2} \frac{x(x+1)\cancel{(x+2)}}{(x-3)\cancel{(x+2)}} \\ &= \lim_{x \to -2} \frac{x(x+1)}{(x-3)} \\ &= \frac{-2\times-1}{-2-3} \\ &= -\frac{2}{5} \end{align}$$

Answer 2

Use the L'Hôpital rule: as you have the indetermination 0/0, derive the functions from the numerator and denominator until your limit has a solution.

Answer 3

L'Hopital's rule states that if $f(n)=g(n)=0$, then $$\lim_{x \to n}\frac{f(x)}{g(x)}=\lim_{x \to n}\frac{f'(x)}{g'(x)}$$ So your limit is the same as $$\lim_{x \to -2}\frac{3x^2+6x+2}{2x-1}$$ Which is $$\frac{3(-2)^2+6(-2)+2}{2(-2)-1}$$ $$\frac{12-12+2}{-4-1}$$ $$-\frac{2}{5}$$

Answer 4

Simplify the fraction first by factoring and canceling the $(x+2)$ term. $$ \begin{align*} &\lim_{x \to -2} \frac{(x+2)(x+1)x}{(x+2)(x-3)} \\ &= \lim_{x \to -2} \frac{(x+1)x}{x-3} \\ &= \frac{(-1)(-2)}{-5} \\ &= \frac{-2}{5} \end{align*} $$

Answer 5

Let us simplify the function

$$f (x)=x\frac {x^2-x-6+4x+8}{x^2-x-6} $$ $$=x (1+4\frac {x+2}{(x+2)(x-3)}) $$ $$=x (1+4\frac {1}{x-3} )$$

the limit is $$-2 (1-\frac {4}{5})=-\frac {2}{5} $$

Answer 6

Hint:

The factorization of the numerator and denominator is elementary as you know one of their roots ($-2$), and the numerator has no independent term.

Then

$$\frac{x^3+3x^2+2x}{x^2-x-6}=\frac{x(x+1)(x+2)}{(x-3)(x+2)}=\frac{x(x+1)}{x-3}$$

for all $x\ne-2$.