<s>[INST] 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 $3^2$.
 [/INST] I realised my problem while typing in the question.
The common limit I was thinking of is $\frac{e^{x}-1}{x} \rightarrow 1$, and not $\frac{e^{nx}-1}{x} \rightarrow 1$, as is the case of $\frac{e^{3x}-1}{x}$.
Thus, I have to change the denominator into $3x$, and the only way of doing so is to add $3$ to the denominator and the numerator. 
The only way of doing so without messing up the beautiful numerator is to add $3^2 = 9$ to the outside, forming $9(\frac{e^{3x}-1}{3x})^2$.
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