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4 votes
4 votes

$\underset{x \to \infty}{\lim} \left( \frac{3x-1}{3x+1} \right) ^{4x}$ equals

  1. $1$
  2. $0$
  3. $e^{-8/3}$
  4. $e^{4/9}$
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4 Answers

5 votes
5 votes

Note: The actual definition of $e$ is below. $$e:=\lim_{x\to \infty}\left(1+\frac{1}{x} \right)^x$$

 

Here, $$\begin{align} \mathrm{L}&=\lim_{x \to \infty} \left(\frac{3x-1}{3x+1}\right)^{4x} \\ &=\lim_{x \to \infty} \left(1-\frac{2}{3x+1}\right)^{4x} \\ &=\lim_{x \to \infty} \left(1-\frac{1}{\frac{3x+1}{2}}\right)^{4x} \end{align}$$

 

Now, Let $y=\frac{3x+1}{2}$. Since $x\to \infty$ hence $y\to \infty$ and $x=\frac{2y-1}{3}$.

$$\begin{align} \therefore \mathrm{L} &=\lim_{y\to \infty}\left( 1-\frac{1}{y} \right)^{4(\frac{2y-1}{3})} \\ &=\lim_{y\to \infty}\left( 1-\frac{1}{y} \right)^{\frac{8y}{3}-\frac{4}{3}} \\ &=\lim_{y\to \infty} \left( \left( 1-\frac{1}{y} \right)^{\frac{8y}{3}}\cdot \left( 1-\frac{1}{y} \right)^{-\frac{4}{3}} \right) \\ &=\lim_{y\to \infty}\left( 1-\frac{1}{y} \right)^{\frac{8y}{3}} \cdot 1 \\ &=\left( \lim_{y\to \infty}\left( 1-\frac{1}{y} \right)^{y} \right)^{\frac{8}{3}} \\ &= \left( e^{-1} \right)^{\frac{8}{3}} \\ &= e^{-\frac{8}{3}} \end{align}$$

So the correct answer is C.

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3 votes
3 votes

Check  if  wrong solution

2 votes
2 votes
$Y=\underset{x \to \infty}{\lim} \bigg( \frac{3x-1}{3x+1} \bigg) ^{4x}$

$\log Y=4\underset{x \to \infty}{\lim} x \log \bigg( \frac{3x-1}{3x+1} \bigg)$

$\log Y=4\underset{x \to \infty}{\lim} \frac{\log \bigg( \frac{3x-1}{3x+1}\bigg)}{1/x} $

$\log Y=4\underset{x \to \infty}{\lim} \frac{\log \bigg( \frac{6}{(3x-1)(3x+1)}\bigg)}{-1/x^{2}} $

         
  $\log Y=8/3$

$Y=e^{8/3}$
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0 votes
0 votes
Answer $C$

Let $ L =\underset{\mathrm{x\to\infty}}{\lim}\Big(\frac{3x-1}{3x+1}\Big)^{4x} = \underset{\mathrm{x\to\infty}}{\lim}\Big(1-\frac{2}{3x+1}\Big)^{4x}  = \underset{\mathrm{x\to\infty}}{\lim}\Big(1-\frac{1}{\frac{3x+1}{2}}\Big)^{4x}$

Now, Let $y = \frac{3x+1}{2}\;\;\because\;\;x \rightarrow\infty\;\therefore y \rightarrow \infty$ and $x = \frac{2y-1}{3}$
So, $$ \therefore L = \underset{\mathrm{y\to\infty}}{\lim}\Big(1-\frac{1}{y}\Big)^{4\Big (\frac{2y-1}{3}\Big )}$$
$$  = \underset{\mathrm{y\to\infty}}{\lim}\Big(1-\frac{1}{y}\Big)^{\frac{8y}{3}-\frac{4}{3}}$$

$$  = \underset{\mathrm{y\to\infty}}{\lim}\Bigg (\Big(1-\frac{1}{y}\Big)^{\frac{2y}{3}}.\Big (1-\frac{1}{y}\Big )^{\frac{-4}{3}}\Bigg )$$

$$  = \underset{\mathrm{y\to\infty}}{\lim}\Bigg (\Big(1-\frac{1}{y}\Big)^{\frac{8y}{3}}.1\Bigg )$$

$$  = \underset{\mathrm{y\to\infty}}{\lim}\Bigg (\Big(1-\frac{1}{y}\Big)^{\frac{8y}{3}}.1\Bigg )$$

$$= \Big (\underset{\mathrm{y\to\infty}}{\lim}\Big({1-\frac{1}{y}\Big )^y}\Big )^{\frac{8}{3}} \qquad \to (1)$$

But, $$ \because \underset{\mathrm{y\to\infty}}{\lim}\Big({1-\frac{1}{y}\Big )^y} = \frac{1}{e} = e^{-1}$$

So, equation $(1)$ reduces to:

$$={(e^{-1})}^{\frac{8}{3}}$$

$$= e^{\frac{-8}{3}}$$

$\therefore \; C$ is the correct option.
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