What is the limit of $\frac{e^{6x}-2e^{3x} + 1}{x^2}$, as $x \rightarrow 0$? I am to calculate $\frac{e^{6x}-2e^{3x} + 1}{x^2}$ when $x$ goes towards $0$.
I find that
$$\frac{e^{6x}-2e^{3x} + 1}{x^2} = \frac{(e^{3x}-1)^2}{x^2} =  \left(\frac{e^{3x}-1}{x}\right)^2$$
$$\left(\frac{e^{3x}-1}{x}\right)^2 \rightarrow 1^2$$
but according to the answer in the book I am incorrect. It agrees with me halfway through, but ends with
$$\frac{(e^{3x}-1)^2}{x^2} =  9\left(\frac{e^{3x}-1}{3x}\right)^2 \rightarrow 9 \times 1^2$$
While this is correct mathematically, why would it be $3$ and $9$ instead of for example $4$ and $16$ or, as in my case, $1$ and $1$? I don't see the relevance of adding the $3$ and $3^2$.

Solution 1 
Since $\lim_{x\to 0} \displaystyle \frac{e^x-1}{x}=1$
$$\lim_{x\to 0}\left(\frac{e^{3x}-1}{3x}\times3\right)^2=9$$
Solution 2
Let $e^{3x}-1=y$ and then the limit turns into
 $$\lim_{x\to 0}\left(\frac{e^{3x}-1}{x}\right)^2=\lim_{y\to 0}\left(\frac{y}{\ln(y+1)}\times3\right)^2=9$$
because $\lim_{y\to0} (1+y)^{1/y}=e$, and then $\lim_{y\to 0} \displaystyle\frac{y}{\ln(y+1)}=1$
