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

This is what I did for:

$$\lim \limits_{x \to \infty} {\left({3x-2 \over3x+4}\right)}^{3x+1}$$

1. Check form: $\left({\infty \over \infty}\right)^{\infty}$.
2. Apply L'Hospital's Rule to just $\lim \limits_{x \to \infty} {\left({3x-2 \over3x+4}\right)}$: $$\lim \limits_{x \to \infty} {3 \over 3} = 1$$
3. So form in step 1, is really $1^{\infty}$.
4. Take the natural log to bring the exponent down: $$\begin{align} \\ & \lim \limits_{x \to \infty} ln\left[{\left({3x-2 \over3x+4}\right)}^{3x+1}\right] \\ &&\\ & = \lim \limits_{x \to \infty} \space (3x+1) \cdot ln\left({3x-2 \over3x+4}\right) \\ \end{align}$$
5. Check form: $\infty \cdot 0$.
6. Rearrange it into quotient form: $$ = \lim \limits_{x \to \infty} \space {ln\left({3x-2 \over3x+4}\right) \over {1 \over 3x+1}}$$
7. Check form: $0 \over 0$
8. Apply L'Hospital's Rule and simplify: $$\begin{align} \\ & = \lim \limits_{x \to \infty} \space {{1 \over {3x-2 \over3x+4}} \cdot {(3x+4)(3)-(3x-2)(3) \over (3x+4)^2} \over {(3x+1)(0)-(1)(3) \over (3x+1)^2}}\\ &&\\ & = \lim \limits_{x \to \infty} \space {{1 \over {3x-2 \over3x+4}} \cdot {18 \over (3x+4)^2} \over {-3 \over (3x+1)^2}}\\ &&\\ & = \lim \limits_{x \to \infty} \space {{18 \over (3x-2)(3x+4)} \over {-3 \over (3x+1)^2}}\\ &&\\ & = \lim \limits_{x \to \infty} \space {18(3x+1)^2 \over -3(3x-2)(3x+4)}\\ &&\\ & = \lim \limits_{x \to \infty} \space -{6(3x+1)^2 \over (3x-2)(3x+4)}\\ \end{align}$$
9. Check form: ${\infty \over {\infty \cdot \infty}} = {\infty \over \infty}$.
10. Apply L'Hospitals Rule again and simplify: $$\begin{align} \\ & =\lim \limits_{x \to \infty} \space -{12(3x+1)(3) \over (3)(3)}\\ &&\\ & =\lim \limits_{x \to \infty} \space -{4(3x+1)}\\ &&\\ & =\lim \limits_{x \to \infty} \space -{4(3x+1)} = \infty\\ \end{align}$$
11. 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?

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The original:

1. Check form: $\left({\infty \over \infty}\right)^{\infty}$.
2. Apply L'Hospital's Rule to just $\lim \limits_{x \to \infty} {\left({3x-2 \over3x+4}\right)}$: $$\lim \limits_{x \to \infty} {3 \over 3} = 1$$
3. So form in step 1, is really $1^{\infty}$.
4. Take the natural log to bring the exponent down: $$\begin{align} \\ & \lim \limits_{x \to \infty} ln\left[{\left({3x-2 \over3x+4}\right)}^{3x+1}\right] \\ &&\\ & = \lim \limits_{x \to \infty} \space (3x+1) \cdot ln\left({3x-2 \over3x+4}\right) \\ \end{align}$$
5. Check form: $\infty \cdot 0$.
6. Rearrange it into quotient form: $$ = \lim \limits_{x \to \infty} \space {ln\left({3x-2 \over3x+4}\right) \over {1 \over 3x+1}}$$
7. Check form: $0 \over 0$
8. Apply L'Hospital's Rule and simplify: $$\begin{align} \\ & = \lim \limits_{x \to \infty} \space {{1 \over {3x-2 \over3x+4}} \cdot {(3x+4)(3)-(3x-2)(3) \over (3x+4)^2} \over {(3x+1)(0)-(1)(3) \over (3x\mathbf{\color{red}{-2}})^2 \tag{ERROR in red}}} \\ &&\\ & = \lim \limits_{x \to \infty} \space {{1 \over {3x-2 \over3x+4}} \cdot {18 \over (3x+4)^2} \over {-3 \over (3x\mathbf{\color{red}{-2}})^2}}\\ &&\\ & = \lim \limits_{x \to \infty} \space {{18 \over (3x-2)(3x+4)} \over {-3 \over (3x\mathbf{\color{red}{-2}})^2}}\\ &&\\ & = \lim \limits_{x \to \infty} \space {18(3x\mathbf{\color{red}{-2}})^2 \over -3(3x-2)(3x+4)}\\ &&\\ & = \lim \limits_{x \to \infty} \space -{6(3x\mathbf{\color{red}{-2}}) \over (3x+4)}\\ \end{align}$$
9. Check form: ${\infty \over \infty}$.
10. Apply L'Hospitals Rule again and simplify: $$\begin{align} \\ & =\lim \limits_{x \to \infty} \space -{6(3) \over (3)}\\ &&\\ & =\lim \limits_{x \to \infty} \space -{6}\\ &&\\ \end{align}$$
11. 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?

Answer 1

You are good until here where you take an incorrect derivative of the denominator.

\begin{align} \lim \limits_{x \to \infty} \space -{6(3x-1)^2 \over (3x-2)(3x+4)} \end{align}

You don't even have to apply L'Hopitals rule here, you can just compare the coefficients of the $x^2$ terms. Don't forget the negative.

$$\lim_{x\to\infty} -\frac{6(3x-1)^2}{(3x-2)(3x+4)} = \lim_{x\to\infty} -\frac{\color{red}{54}x^2-36x+6}{\color{red}{9}x^2+6x-8} = -\frac{54}{9} = -6$$

Answer 2

Starting with

$$\lim_{x\to\infty}\left(\frac{3x-2}{3x+4}\right)^{3x+1}$$

we modify it to

$$\lim_{x\to\infty}\left(1+\frac{-6}{3x+4}\right)^{3x+1}=\lim_{x\to\infty}\left(1-\frac{6}{3x+4}\right)^{3x+4}\cdot\left(1-\frac 6{3x+4}\right)^{-3}$$

which immediately reduces by $y=3x+4$ to

$$\lim_{y\to\infty}\left(1-\frac 6y\right)^y\cdot\left(1-\frac 6y\right)^{-3}=e^{-6}\cdot 1^{-3}=e^{-6}$$

Answer 3

Note that the derivative of the denominator $(3x-2)(3x+4)$ is not $(3)\cdot (3)$, but something else. There is no case of L'Hopital of $\frac{\infty}{\infty\cdot\infty}$, only a case of $\frac{\infty}{\infty}$.

Alternative answer. Write it as:

$$\lim_{x\to\infty} \left(1-\frac{6}{3x+4}\right)^{3x+1}$$

Let $z=\frac{3x+4}{6}$ Then $3x+1=6z-3$ so this is:

$$\lim_{z\to\infty} \left(1-\frac{1}{z}\right)^{6z-3}$$

Now, $\left(1-\frac{1}{z}\right)^z\to e^{-1}$ and $\left(1-\frac{1}{z}\right)^{-3}\to 1$. So the limit is $e^{-6}$.

Answer 4

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 $1^{-3}=1$.