TITLE: I got the answer for $\lim \limits_{x \to \infty} {\left({3x-2 \over3x+4}\right)}^{3x+1}$, but only by a mistake - how do I solve correctly?
QUESTION [14 upvotes]: This is what I did for:
$$\lim \limits_{x \to \infty} {\left({3x-2 \over3x+4}\right)}^{3x+1}$$

Check form: $\left({\infty \over \infty}\right)^{\infty}$.
Apply L'Hospital's Rule to just $\lim \limits_{x \to \infty} {\left({3x-2 \over3x+4}\right)}$:
$$\lim \limits_{x \to \infty} {3 \over 3} = 1$$
So form in step 1, is really $1^{\infty}$.
Take the natural log to bring the exponent down:
$$\begin{align} \\
& \lim \limits_{x \to \infty} ln\left[{\left({3x-2 \over3x+4}\right)}^{3x+1}\right] \\
&&\\
& = \lim \limits_{x \to \infty} \space (3x+1) \cdot ln\left({3x-2 \over3x+4}\right) \\
\end{align}$$
Check form: $\infty \cdot 0$.
Rearrange it into quotient form:
$$ = \lim \limits_{x \to \infty} \space {ln\left({3x-2 \over3x+4}\right) \over {1 \over 3x+1}}$$
Check form: $0 \over 0$
Apply L'Hospital's Rule and simplify:
$$\begin{align} \\
& = \lim \limits_{x \to \infty} \space {{1 \over {3x-2 \over3x+4}} \cdot {(3x+4)(3)-(3x-2)(3) \over (3x+4)^2} \over {(3x+1)(0)-(1)(3) \over (3x+1)^2}}\\
&&\\
& = \lim \limits_{x \to \infty} \space {{1 \over {3x-2 \over3x+4}} \cdot {18 \over (3x+4)^2} \over {-3 \over (3x+1)^2}}\\
&&\\
& = \lim \limits_{x \to \infty} \space {{18 \over (3x-2)(3x+4)} \over {-3 \over (3x+1)^2}}\\
&&\\
& = \lim \limits_{x \to \infty} \space {18(3x+1)^2 \over -3(3x-2)(3x+4)}\\
&&\\
& = \lim \limits_{x \to \infty} \space -{6(3x+1)^2 \over (3x-2)(3x+4)}\\
\end{align}$$
Check form: ${\infty \over {\infty \cdot \infty}} = {\infty \over \infty}$.
Apply L'Hospitals Rule again and simplify:
$$\begin{align} \\
& =\lim \limits_{x \to \infty} \space -{12(3x+1)(3) \over (3)(3)}\\
&&\\
& =\lim \limits_{x \to \infty} \space -{4(3x+1)}\\
&&\\
& =\lim \limits_{x \to \infty} \space -{4(3x+1)} = \infty\\
\end{align}$$
This implies $\lim \limits_{x \to \infty} {\left({3x-2 \over3x+4}\right)}^{3x+1} = e^{\infty} = \infty$. 

But this is wrong, the above work is from me double checking my work and finding an error. I did have the correct answer originally (not in the above work), but with the wrong work to find it. Can anyone help me figure out where I went wrong in the original work and where I went wrong above?

The original:

Check form: $\left({\infty \over \infty}\right)^{\infty}$.
Apply L'Hospital's Rule to just $\lim \limits_{x \to \infty} {\left({3x-2 \over3x+4}\right)}$:
$$\lim \limits_{x \to \infty} {3 \over 3} = 1$$
So form in step 1, is really $1^{\infty}$.
Take the natural log to bring the exponent down:
$$\begin{align} \\
& \lim \limits_{x \to \infty} ln\left[{\left({3x-2 \over3x+4}\right)}^{3x+1}\right] \\
&&\\
& = \lim \limits_{x \to \infty} \space (3x+1) \cdot ln\left({3x-2 \over3x+4}\right) \\
\end{align}$$
Check form: $\infty \cdot 0$.
Rearrange it into quotient form:
$$ = \lim \limits_{x \to \infty} \space {ln\left({3x-2 \over3x+4}\right) \over {1 \over 3x+1}}$$
Check form: $0 \over 0$
Apply L'Hospital's Rule and simplify:
$$\begin{align} \\
& = \lim \limits_{x \to \infty} \space {{1 \over {3x-2 \over3x+4}} \cdot {(3x+4)(3)-(3x-2)(3) \over (3x+4)^2} \over {(3x+1)(0)-(1)(3) \over (3x\mathbf{\color{red}{-2}})^2 \tag{ERROR in red}}} \\
&&\\
& = \lim \limits_{x \to \infty} \space {{1 \over {3x-2 \over3x+4}} \cdot {18 \over (3x+4)^2} \over {-3 \over (3x\mathbf{\color{red}{-2}})^2}}\\
&&\\
& = \lim \limits_{x \to \infty} \space {{18 \over (3x-2)(3x+4)} \over {-3 \over (3x\mathbf{\color{red}{-2}})^2}}\\
&&\\
& = \lim \limits_{x \to \infty} \space {18(3x\mathbf{\color{red}{-2}})^2 \over -3(3x-2)(3x+4)}\\
&&\\
& = \lim \limits_{x \to \infty} \space -{6(3x\mathbf{\color{red}{-2}}) \over (3x+4)}\\
\end{align}$$
Check form: ${\infty \over \infty}$.
Apply L'Hospitals Rule again and simplify:
$$\begin{align} \\
& =\lim \limits_{x \to \infty} \space -{6(3) \over (3)}\\
&&\\
& =\lim \limits_{x \to \infty} \space -{6}\\
&&\\
\end{align}$$
This then works out to $\lim \limits_{x \to \infty} {\left({3x-2 \over3x+4}\right)}^{3x+1} = e^{-6}$

$e^{-6}$ is the correct answer, but I thought there was probably a better way to solve this then the approach I took so I reviewed it and then found the error. Me stumbling onto the right answer was an accident. When I tried to fix the error I get the wrong answer. Can anyone make sense of this for me and show me what I'm doing wrong, and how to solve this problem correctly?

REPLY [3 votes]: Here's how I'd tackle the problem: this way has the advantage of not needing L'Hopital at all (as long as you know the classic limit definition of $e$).  We have a denominator of $3x+4$, which is a really inconvenient value, so let's make a change of variables and set $y=3x+4$; then $\lim\limits_{x\to\infty}$ translates to $\lim\limits_{y\to\infty}$, and our expression becomes $\lim\limits_{y\to\infty}\left(\dfrac{y-6}{y}\right)^{y-3}$.  Now we can simplify the expression in parenthesis: $\lim\limits_{y\to\infty}\left(1+\dfrac{(-6)}{y}\right)^{y-3}$ = $\lim\limits_{y\to\infty}\left(1+\dfrac{(-6)}{y}\right)^{y}\cdot\left(1+\dfrac{(-6)}{y}\right)^{-3}$.  By using the standard limit $\lim\limits_{n\to\infty}\left (1+\frac xn\right)^n = e^x$ the left term evaluates to $e^{-6}$, and of course the right term goes to $1^{-3}=1$.