Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $3^2$.

share|improve this question
3  
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
add comment

4 Answers 4

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$$\lim_{x\to 0} \frac{f(x)-f(0)}{x} = f'(0)=3e^{0}=3$$

share|improve this answer
add comment

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 $9(\frac{e^{3x}-1}{3x})^2$.

share|improve this answer
add comment

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

share|improve this answer
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.