# Thread: Limit of a function

1. ## Limit of a function

I've got a problem with the following limit..

$\displaystyle (x -> 0) lim [(x^3 - 9x^2 + 27x - 27) / (x^2 - x -6)]$
can someone solve this.. the solution is 0 but i dunno how.. i get 9/2

2. Originally Posted by metlx
$\displaystyle (x->0) lim (x3 - 9x2 + 27x - 27) / (x2 - x - 6)$
(btw, how do you use the math tags properly? so the 'x->0' would go below 'lim' and 'x2' would have a uppercase 2)
\lim_{x \to 0} (x^3 - 9x^2 + 27x - 27) / (x^2 - x - 6)

$\displaystyle \lim _{x \to 0} (x^3 - 9x^2 + 27x - 27) / (x^2 - x - 6)$

3. Hello, metlx!

I don't agree with their answer either,
. . but I bet I know where the error is . . .

$\displaystyle \lim_{x\to0}\frac{x^3 - 9x^2 + 27x - 27}{x^2-x-6}$

As it is written, we can simply substitute $\displaystyle x = 0$

. . and we get: .$\displaystyle \frac{0-0+0-27}{0-0-7} \:=\:\frac{-27}{-6} \;=\;\frac{9}{2}$

I'm certain the limit is: .$\displaystyle {\color{red}x\to3}$

. . $\displaystyle \lim_{x\to3}\frac{x^3-9x^2+27x-27}{x^2-x-6} \:=\:\frac{0}{0}$ . . . an indeterminate form.

We have: .$\displaystyle \frac{x^3-9x^2+27x-27}{x^2-x-6} \:=\:\frac{(x-3)^3}{(x-3)(x+2)}$

. . Since $\displaystyle x \neq 3$, we can reduce: .$\displaystyle \frac{(x-3)^2}{x+2}$

Therefore: .$\displaystyle \lim_{x\to3}\frac{(x-3)^2}{x+2} \;=\;\frac{0^2}{5} \;=\;0$

4. thanks!

I've got another problem with roots..

$\displaystyle \lim_{x\to4}\frac{\sqrt{1+2x}-3}{\sqrt{x}-2}$

the answer is 4/3.. but how?

5. never mind. i managed to solve it =)

$\displaystyle \lim_{x\to4}\frac{\sqrt{1+2x}-3}{\sqrt{x}-2} = \lim_{x\to4}\frac{(1+2x-9) (\sqrt{x}+2)}{(x-4) (\sqrt{1+2x}+3)}= \lim_{x\to4}\frac{ 4(2x-8)}{6(x-4)} = \lim_{x\to4}\frac{8(x-4)}{6(x-4)} = \frac {4}{3}$