I've been practicing limits today, and I've came across this exercise which confused me a bit:
$\lim_{x\to -3} \frac{(x^2-9)^2}{(x+3)^2}$
My approach was to do this:
$\lim_{x\to -3} \frac{(x^2-9)^2}{x^2+3x+9}$
$\lim_{x\to -3} \frac{0}{9}$
Can anyone explain why this is not a valid approach? I realised that as long as the denominator is not 0, I can start inserting the x.
The valid solution is 36, and I know how to get there, I just don't get it why my approach is wrong.
Thanks!
On
$$\begin{align} x^2 - 9 &= x^2 - 3^2 \\ \\ &= (x+3)(x-3) \\ \Rightarrow \dfrac{(x^2-9)^2}{(x+3)^2} &= \dfrac{\require{cancel}\cancel{(x+3)^2}(x-3)^2}{\cancel{(x+3)^2}} \\ &= (x-3)^2. \\ \\ \therefore \lim_{x\to -3}\frac{(x^2-9)^2}{(x+3)^2} &= \lim_{x\to -3} (x-3)^2 \\ \\ &= (-3 - 3)^2 \\ \\ &= 36.\end{align}$$
Note that
$$\lim_{x\to -3} \frac{(x^2-9)^2}{x^2+\color{red}6x+9}$$
We can solve in this way
$$\lim_{x\to -3} \frac{(x^2-9)^2}{(x+3)^2}=\lim_{x\to -3} \frac{(x-3)^2(x+3)^2}{(x+3)^2}=\lim_{x\to -3} (x-3)^2=36$$