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${\lim_{x\to \infty} (\frac{3x-1}{3x+1})^{4x}}$ = ?

P.S. - I tried reducing it to some form like $\lim_{n\to \infty} (1 - \frac1n)^n$, the value of which is $e^{-1}$. Was I correct?

Pedro
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A.Chakraborty
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    Something like that works (when you do it right). – Daniel Fischer Apr 23 '14 at 19:49
  • To continue Daniel Fischer's hint, use long division to rewrite $\frac{3x-1}{3x+1}$ and then consider the variable change $u = 3x + 1$ (or alternatively, express $4x$ as a constant times $3x+1$ plus another constant). – Dave L. Renfro Apr 23 '14 at 19:59

3 Answers3

5

Your idea is good and just you need some algebra:

$$\large\left(\frac{3x-1}{3x+1}\right)^{4x}=\left(1-\frac{1}{\frac{3x+1}2}\right)^{4x}=\left(\underbrace{\left(1-\frac{1}{\frac{3x+1}2}\right)^{\frac{3x+1}2}}_{\xrightarrow{x\to\infty} e^{-1}}\right)^{\frac{4x}{\frac{3x+1}2}}\xrightarrow{x\to\infty}e^{-\frac83}$$

2

Hint: Since all values are positive for large enough $x$,

$$ \lim_{x\rightarrow\infty}\left(\frac{3x-1}{3x+1}\right)^{4x} = \lim_{x\rightarrow\infty}\exp\left(\ln\left(\left(\frac{3x-1}{3x+1}\right)^{4x}\right)\right) = \exp\left(\lim_{x\rightarrow\infty}\ln\left(\left(\frac{3x-1}{3x+1}\right)^{4x}\right)\right) = \exp\left(\lim_{x\rightarrow\infty}4x\left(\ln\left(\frac{3x-1}{3x+1}\right)\right)\right). $$ Now you can use L'Hospital's rule on the inner limit.

MartianInvader
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As the indetermination is of the form $1^{\infty}$, use the exponential fundamental limit $\displaystyle\lim_{v\rightarrow \infty}(1+v)^{\frac{1}{v}}=e$:

$$ \lim_{x\rightarrow \infty}\left(\frac{3x-1}{3x+1}\right)^{4x}=\lim_{x\rightarrow \infty}\left(1+\frac{3x-1}{3x+1}-1\right)^{4x}=\lim_{x\rightarrow \infty}\left(1+\frac{-2}{3x+1}\right)^{4x}=$$ $$=\lim_{x\rightarrow \infty}\left(1+\frac{-2}{3x+1}\right)^{\frac{3x+1}{-2}\cdot \frac{-8x}{3x+1}}= e^{\lim_{x\rightarrow \infty}\frac{-8x}{3x+1}}=e^{-\frac{8}{3}}.$$

ZHN
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