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

1. $1$
2. $0$
3. $e^{-8/3}$
4. $e^{4/9}$
in Calculus
recategorized | 35 views

+1 vote
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 (19.2k points)
0

@JEET Why is this approach wrong ?

if we divide numerator and denominator by x

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

and now subsititue x as $\infty$

$\lim_{x\rightarrow \infty}\left ( \frac{3}{3} \right )^{\infty}$

$\lim_{x\rightarrow \infty}\left ( 1 \right )^{\infty}$ = 1

0
Because this is not defined.

$\because$ You don't know how many times you have to multiply 1.

The question was given for making the student fall into this trap only.
0
0
" You don't know how many times you have to multiply 1"

What do you mean by this? In the link also they have written if in $\lim_{x\rightarrow \infty}(1)^{x}$ if 1 is absolute then answer is 1, but if in $\lim_{x\rightarrow \infty}( approx1)^{x}$ then limit is indeterminant.

Also, check this approach.

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

When we put x as $\infty$, it actually means that we put very large value for x.

So in actual we get

$\lim_{x\rightarrow \infty}\left ( \frac{3-0.00..01}{3+0.00..01} \right )^{4x}$

$\lim_{x\rightarrow \infty}\left ( \frac{2.999..9}{3.00..01} \right )^{4x}$

$\lim_{x\rightarrow \infty}\left ( <1 \right )^{4x}$ = 0

0

@JEET See this comment from the link you shared.

+1
Then you yourself said it in the first line.

So, $(approx \;1) ^{4x}$ when $x \to \infty$ is not correct.

What I meant by my comment is that I tried to give the intuition of why $1^{\infty}$ is indeterminate form.

This can be very debatable and long.

You can read a lot of articles, probably on the stackexchange.