150 views

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

1. $1$
2. $0$
3. $e^{-8/3}$
4. $e^{4/9}$
in Calculus
recategorized | 150 views
0
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 $y = \frac{3x+1}{2}$. Since, $x \rightarrow\infty\;\therefore y \rightarrow \infty$ 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.
0

@`JEET

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

+1
Thanks for making this point clear.

Actually I was confused in this, as many places $0.\infty = 0$ is mentioned.
0
can you please give link where $0.∞=0$ is mentioned
0
Ok. Let me search again.
0

@ankitgupta.1729

Is indeterminate form applicable here?

0
what is intermediate form ?
+1

Credit goes to for making the correctons in the answer.

0

@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..

+1
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.

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.

by Active (3.5k points)
edited
0
yes, but indeterminate form, not applicable.

right?
0
Why not?

There is no doubt that the answer is $e^{-8/3}$. If you want to have some experiments, just put the values of $x=5,10,50,100,1000, ...$ to the expression $\left( \frac{3x-1}{3x+1} \right)^{4x}$, you'll notice the value the expression gets (converges) closer to $e^{-8/3}=0.069483451....$

Another Note:
$$e:=\lim_{x\to \infty} \left( 1+\frac{1}{x} \right)^x$$

Using generalized binomial expansion it can be reduced to $$e=\sum_{n=0}^{\infty}\frac{1}{n!}=2.718281828...$$
0

How L'hospital applicable here?? not possible I think.
0

Yes. $$\lim_{x\to \infty} \left( \frac{3x-1}{3x+1} \right)^{4x}=\lim_{x\to \infty} \frac{(3x-1)^{4x}}{(3x+1)^{4x}}=\lim_{x\to \infty} \frac{f(x)}{g(x)} ~~[\mathrm{Let}]$$

Now, $f'(x)=\left\{ 4\ln(3x-1) +\frac{12x}{3x-1} \right\}(3x-1)^{4x}\\ g'(x)=\left\{ 4\ln(3x+1) +\frac{12x}{3x+1} \right\}(3x+1)^{4x}$.

Here we can notice that both functions have exponential part aka $(3x\pm1)^{4x}$ respectively which always remain even if we further take derivatives. So L' Hospital rule doesn't work here like this.

Even if you take the form as $$\lim_{x\to \infty} \left( \frac{3x-1}{3x+1} \right)^{4x}=\lim_{x\to \infty} \left( \frac{3-\frac{1}{x}}{3+\frac{1}{x}} \right)^{4x}=\lim_{\mathrm{Let~} y=\frac{1}{x}\\ \Rightarrow y\to 0} \frac{(3-y)^{\frac{4}{y}}}{(3+y)^{\frac{4}{y}}}=\lim_{y\to 0} \frac{f(y)}{g(y)} ~~[\mathrm{Let}]$$

L' Hospital rule still doesn't work like this.

0
Is there any alternative solution?
0

@techbd123

Which formula you used in the last second line?

0

$\displaystyle \lim_{y \to \infty} \left(1-\frac{1}{y}\right)^y = \lim_{\mathrm{Let~} u=-y\\ y \to \infty\\ \Rightarrow \frac{1}{y} \to 0 \Rightarrow -\frac{1}{u} \to 0 \\ \Rightarrow \frac{1}{u} \to 0 } \left(1+\frac{1}{u}\right)^{(-u)} =\left( \lim_{\frac{1}{u} \to 0} \left(1+\frac{1}{u}\right)^u \right)^{-1}=e^{-1}$

Yes. It's using generalized binomial expansion which is straight forward but a bit tiresome.

$\displaystyle \lim_{x \to \infty} \left( \frac{3x-1}{3x+1} \right)^{4x}=\lim_{x \to \infty} \left(1- \frac{2}{3x+1} \right)^{4x}\\=\displaystyle \lim_{x \to \infty}\left[ 1-\frac{4x}{1!} \left(\frac{2}{3x+1} \right)+ \frac{4x(4x-1)}{2!} \left(\frac{2}{3x+1} \right)^2 - \frac{4x(4x-1)(4x-2)}{3!} \left(\frac{2}{3x+1} \right)^3+ \cdots \right] \\ \scriptsize \displaystyle =\lim_{x \to \infty}\left[ 1-\frac{4x}{1!} \frac{1}{x}\left(\frac{2}{3+\frac{1}{x}} \right)+ \frac{4x^2(4-\frac{1}{x})}{2!} \frac{1}{x^2} \left(\frac{2}{3+\frac{1}{x}} \right)^2 - \frac{4x^3(4-\frac{1}{x})(4-\frac{2}{x})}{3!} \frac{1}{x^3} \left(\frac{2}{3+\frac{1}{x}} \right)^3+ \cdots \right] \\ \small \displaystyle =\lim_{x \to \infty\\ \Rightarrow \frac{1}{x} \to 0} \left[ 1-\frac{4}{1!}\left(\frac{2}{3+\frac{1}{x}} \right)+ \frac{4(4-\frac{1}{x})}{2!}\left(\frac{2}{3+\frac{1}{x}} \right)^2 - \frac{4(4-\frac{1}{x})(4-\frac{2}{x})}{3!} \left(\frac{2}{3+\frac{1}{x}} \right)^3+ \cdots \right] \\ =1-\frac{4}{1!}\left(\frac{2}{3} \right)+ \frac{4^2 }{2!}\left(\frac{2}{3} \right)^2 - \frac{4^3}{3!} \left(\frac{2}{3} \right)^3+ \cdots \\ =1+\frac{\left(-\frac{8}{3} \right)}{1!}+ \frac{\left(-\frac{8}{3} \right)^2 }{2!} + \frac{\left(-\frac{8}{3} \right)^3}{3!} + \cdots \\= e^{-\frac{8}{3}}$

0
Yes, I got it.

Thanks
+1

@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

0
See below, I tried but giving wrong result
+1

@ankitgupta.1729

You are right.

I provided the full solution here.

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

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

Check  if  wrong solution

by Junior (775 points)
0
If you are dividing by x then you have to divide $4x$ by $x$ as well.
0
Your calculation part is not correct.
0

@amit166

which formula r u using??

+1

1 ki power infinity for limit calculation

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

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

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

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

$\log Y=8/3$

$Y=e^{8/3}$
by Veteran (119k points)
edited by
0

It is case of $\large 1^{\infty }$ so answers is $\large e^{-8/3 }$

0
How??
+2
$logY = 4*\lim_{x\rightarrow \infty } x* ln\left ( \frac{3x-1}{3x+1} \right )$

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

Now, RHS is in 0/0 form. Now, can you proceed further ?
+1
Thanks
0

@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 $C$

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.
by Boss (18.9k points)
edited by