<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] Another approach to write $$\frac{e^{3x}-1}{x}=\frac{e^{3x}-e^{3\cdot 0}}{x}=\frac{f(x)-f(0)}{x}$$
Where $f(x)=e^{3x}$ then$$\lim_{x\to 0} \frac{f(x)-f(0)}{x} = f'(0)=3e^{0}=3$$
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