Functions Question 149
Question: $ \underset{x\to \infty }{\mathop{\lim }},{{( \frac{x+2}{x+1} )}^{x+3}} $ is
[MNR 1994]
Options:
A) $ 1 $
B) $ e $
C) $ e^{2} $
D) $ e^{3} $
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ A=\underset{x\to \infty }{\mathop{\lim }}{{( \frac{x+2}{x+1} )}^{x+3}} $ $ =\underset{x\to \infty }{\mathop{\lim }}{{( 1+\frac{1}{x+1} )}^{x+3}} $ $ =\underset{x\to \infty }{\mathop{\lim }}{{[ {{( 1+\frac{1}{x+1} )}^{x+1}} ]}^{\frac{,(x+3)}{(x+1)}}}=e $ $ { \because \underset{x\to \infty }{\mathop{\lim }}{{( 1+\frac{1}{x+1} )}^{x+1}}=e . $ and $ \underset{x\to \infty }{\mathop{\lim }},\frac{,(x+3)}{(x+1)}=. \underset{x\to \infty }{\mathop{\lim }},\frac{,{ 1+(3/x) }}{{ 1+(1/x) }}=1 } $ .
BETA
