Functions Question 291
Question: The value of $ \underset{x\to -1}{\mathop{\lim }},\frac{x^{2}+3x+2}{x^{2}+4x+3} $ is equal to
[Pb. CET 2000]
Options:
A) 0
B) 1
C) 2
D) ½
Show Answer
Answer:
Correct Answer: D
Solution:
$ \underset{x\to 1}{\mathop{\lim }},\frac{x^{2}+3x+2}{x^{2}+4x+3}=\underset{x\to -1}{\mathop{\lim }},\frac{x^{2}+2x+x+2}{x^{2}+3x+x+3} $ $ =\underset{x\to -1}{\mathop{\lim }},\frac{(x+1)(x+2)}{(x+1)(x+3)}=\underset{x\to -1}{\mathop{\lim }},\frac{x+2}{x+3}=\frac{1}{2} $ .
BETA
