Functions Question 175
Question: If $ f(x)=\frac{2}{x-3},\ g(x)=\frac{x-3}{x+4} $ and $ h(x)=-\frac{2(2x+1)}{x^{2}+x-12}, $ then $ \underset{x\to 3}{\mathop{\lim }},[f(x)+g(x)+h(x)] $ is
Options:
A) $ -2 $
B) $ -1 $
C) $ -\frac{2}{7} $
D) 0
Show Answer
Answer:
Correct Answer: C
Solution:
We have $ f(x)+g(x)+h(x)=\frac{x^{2}-4x+17-4x-2}{x^{2}+x-12} $ $ =\frac{x^{2}-8x+15}{x^{2}+x-12}=\frac{(x-3),(x-5)}{(x-3),(x+4)} $
$ \therefore \underset{x\to 3}{\mathop{\lim }}[f(x)+g(x)+h(x)]=\underset{x\to 3}{\mathop{\lim }},\frac{(x-3)(x-5)}{(x-3)(x+4)}=-\frac{2}{7} $ .
BETA
