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KCET 2004 Math Solved Paper

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Question : 11 of 60

Marks: +1, -0

\lim\limits_{x \rightarrow\infty}\left(1-\frac{4}{x-1}\right)^{3x-1}=

$e^{12}$
. $e^{-12}$
$e^4$
$e^3$

Solution:

Let

t=x-1

, so

t\to\infty

as

x\to\infty

.

Then

x=t+1

and

3x-1=3(t+1)-1=3t+2

.

Thus the limit becomes

\lim\limits_{t\to\infty}\left(1-\frac{4}{t}\right)^{3t+2}

.

Rewrite it as

\lim\limits_{t\to\infty}\left[\left(1-\frac{4}{t}\right)^t\right]^3\left(1-\frac{4}{t}\right)^2

.

Since

\lim\limits_{t\to\infty}\left(1-\frac{4}{t}\right)^t=e^{-4}

and

\lim\limits_{t\to\infty}\left(1-\frac{4}{t}\right)^2=1

,

the limit is

(e^{-4})^3\cdot 1=e^{-12}

.

Therefore the correct option is

e^{-12}

.

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