A) 0
B) 1
C) 2
D) ½
Correct Answer: D
Solution :
\[\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{2}}+3x+2}{{{x}^{2}}+4x+3}=\underset{x\to -1}{\mathop{\lim }}\,\frac{{{x}^{2}}+2x+x+2}{{{x}^{2}}+3x+x+3}\] \[=\underset{x\to -1}{\mathop{\lim }}\,\frac{(x+1)(x+2)}{(x+1)(x+3)}=\underset{x\to -1}{\mathop{\lim }}\,\frac{x+2}{x+3}=\frac{1}{2}\].You need to login to perform this action.
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