Functions Question 294
Question: The value of $ \underset{x\to \infty }{\mathop{\lim }},{{( \frac{3x-4}{3x+2} )}^{\frac{x+1}{3}}} $ is equal to
[Pb. CET 2004]
Options:
A) $ {e^{-1/3}} $
B) $ {e^{-2/3}} $
C) $ {e^{-1}} $
D) $ {e^{-2}} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \underset{x\to \infty }{\mathop{\lim }},{{( \frac{3x-4}{3x+2} )}^{\frac{x+1}{3}}}=\underset{x\to \infty }{\mathop{\lim }},{{( \frac{3x+2-6}{3x+2} )}^{\frac{x+1}{3}}} $ $ =\underset{x\to \infty }{\mathop{\lim }},{{( 1-\frac{6}{3x+2} )}^{\frac{x+1}{3}}}=\underset{x\to \infty }{\mathop{\lim }},{{[ {{( 1-\frac{6}{3x+2} )}^{\frac{3x+2}{-6}}} ]}^{\frac{-6}{3x+2}.\frac{x+1}{3}}} $ $ =\underset{x\to \infty }{\mathop{\lim }},{e^{\frac{-2(x+1)}{3x+2}}}={e^{-2/3}} $ , $ { \because \underset{x\to \infty }{\mathop{\lim }},\frac{-2(x+1)}{3x+2}=\frac{-2}{3} } $ .
BETA
