Thread: Stuck on easy limit

Hi all, stuck on the following limit:

$\displaystyle \lim_{x \to 2} \frac{t^3+3t^2-12t+4}{t^3-4t}$

I have managed to factor it out to:

$\displaystyle \lim_{x \to 2} \frac{(x-\frac{\sqrt{33}}{2}-\frac{5}{2})(x+\frac{\sqrt{33}}{2}-\frac{5}{2})}{x(x-2)}, x \neq 2 $

Thanks

Originally Posted by Oiler

Hi all, stuck on the following limit:

$\displaystyle \lim_{x \to 2} \frac{t^3+3t^2-12t+4}{t^3-4t}$

I have managed to factor it out to:

$\displaystyle \lim_{x \to 2} \frac{(x-\frac{\sqrt{33}}{2}-\frac{5}{2})(x+\frac{\sqrt{33}}{2}-\frac{5}{2})}{x(x-2)}, x \neq 2 $

Thanks

$\displaystyle \dfrac{t^3 + 3t^2 - 12t + 4}{t^3 - 4t} = \dfrac{(t-2)(t^2+5t-2)}{t(t-2)(t+2)}$

Originally Posted by Oiler

Hi all, stuck on the following limit:

$\displaystyle \lim_{x \to 2} \frac{t^3+3t^2-12t+4}{t^3-4t}$

I have managed to factor it out to:

$\displaystyle \lim_{x \to 2} \frac{(x-\frac{\sqrt{33}}{2}-\frac{5}{2})(x+\frac{\sqrt{33}}{2}-\frac{5}{2})}{x(x-2)}, x \neq 2 $

Thanks

You could try

$\displaystyle \displaystyle\frac{t^3+3t^2-12t+4}{t^3-4t}=\frac{\left(t^3-4t\right)+3t^2-8t+4}{t^3-4t}=1+\frac{3t^2-8t+4}{t^3-4t}$

$\displaystyle =\displaystyle\ 1+\frac{(3t-2)(t-2)}{t\left(t^2-4\right)}=1+\frac{(3t-2)(t-2)}{t(t+2)(t-2)}$