Functions Question 271
Question: $ \underset{x\to \infty }{\mathop{\lim }},{{( \frac{x+3}{x+1} )}^{x+1}}= $
[RPET 2003, UPSEAT 2003]
Options:
A) $ e^{2} $
B) $ e^{3} $
C) e
D) $ {e^{-1}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \underset{x\to \infty }{\mathop{\lim }},{{( \frac{x+3}{x+1} )}^{x+1}} $ $ =\underset{x\to \infty }{\mathop{\lim }},{{( 1+\frac{2}{x+1} )}^{\frac{x+1}{2}.2}} $ $ ={{{ \underset{x\to \infty }{\mathop{\lim }},{{( 1+\frac{2}{x+1} )}^{\frac{x+1}{2}}} }}^{2}} $ $ =e^{2} $ .
BETA
