# 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 $3^2$.

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Hint: The book tries to make the exponent of $e$ equal to the factor of $x$ in the denominator. – Anon Jan 20 '13 at 23:11

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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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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You can also use L'Hôpital's rule: $$\lim_{x\to 0}\frac{e^{3x}-1}{x}=\lim_{x\to 0}\frac{3e^{3x}}{1}=3.$$

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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 $\lim_{y\to 0} \displaystyle\frac{y}{\ln(y+1)}=1$

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