ISI2014-DCG-3

Question

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

• A. $1$
• B. $0$
• C. $e^{-8/3}$
• D. $e^{4/9}$

Comments

Credit goes to @techbd123 for making the correctons in the answer.

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@ankitgupta.1729

oh yes, it will be indeterminate right?

But is L hospital applicable here?like this

$0*\infty$ is an indeterminate form..it is not zero..

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yeah, you can use L'hospital rule here. Think how to convert $1^{\infty}$ form to $\infty/\infty$ form..think about log both sides and proceed further.

Answer 1

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.

Comments

@srestha

You can solve this question by using L'Hopital's rule also :)

Take log  both sides and convert new expression of right hand side into 0/0 form

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See below, I tried but giving wrong result

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@ankitgupta.1729

You are right.

I provided the full solution here.

Let $\displaystyle L = \lim_{x \to \infty} \left( \frac{3x-1}{3x+1} \right)^{4x}\\ \displaystyle  \Rightarrow \ln L = \lim_{x \to \infty} 4x\ln\left( \frac{3x-1}{3x+1} \right) \\ \displaystyle = 4\lim_{\mathrm{Let~} y=\frac{1}{x}\\ \Rightarrow y \to 0} \frac{\ln\left( \frac{\frac{3}{y}-1}{\frac{3}{y}+1} \right)}{y}=4\lim_{y \to 0} \frac{\ln\left( \frac{3-y}{3+y} \right)}{y}\\ \displaystyle=4\lim_{y \to 0} \frac{\ln(3-y)-\ln(3+y)}{y}=4\lim_{y \to 0} \frac{\frac{-1}{3-y}-\frac{1}{3+y}}{1}; ~[\mathrm{Using~L'~Hospital~Rule}] \\ \displaystyle =4\left(-\frac{1}{3}-\frac{1}{3}\right)=-\frac{8}{3}$

 

$\displaystyle \therefore \ln L =-\frac{8}{3}\Rightarrow L=e^{-\frac{8}{3}}$

Answer 2

Check  if  wrong solution

Comments

If you are dividing by x then you have to divide $4x$ by $x$ as well.

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Your calculation part is not correct.

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@amit166

which formula r u using??

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https://math.stackexchange.com/questions/1987215/1-to-the-power-of-infinity-formula 

1 ki power infinity for limit calculation

Answer 3

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

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

$\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}} $

         
  $\log Y=8/3$

$Y=e^{8/3}$

Comments

$logY = 4*\lim_{x\rightarrow \infty } x* ln\left ( \frac{3x-1}{3x+1} \right )$

$\Rightarrow$ $logY = 4*\lim_{x\rightarrow \infty }  ln\left ( \frac{3x-1}{3x+1} \right )/(\frac{1}{x})$

Now, RHS is in 0/0 form. Now, can you proceed further ?

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Thanks

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@srestha

There are some misktakes. Please check the equations again.

It will be
$\frac{\mathrm{d}}{\mathrm{d}x}\log\left(\frac{3x-1}{3x+1}\right)=\frac{6}{(3x-1)(3x+1)}$
then evaluating the rest will produce $-\frac{8}{3}$ not $\frac{8}{3}$

Answer 4

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.