Functions Question 203
Question: The value of $ \underset{x\to -\infty }{\mathop{\lim }},\frac{\sqrt{4x^{2}+5x+8}}{4x+5} $ is
[Roorkee 1998]
Options:
A) $ -1/2 $
B) 0
C) $ 1/2 $
D) 1
Show Answer
Answer:
Correct Answer: A
Solution:
$ \underset{x,\to ,-\infty }{\mathop{\lim }}\frac{\sqrt{4x^{2}+5x+8}}{4x+5} $ $ =\underset{h\to 0}{\mathop{\lim }},\frac{\sqrt{4,{{(-1/h)}^{2}}+5,(-1/h)+8}}{4,(-1/h)+5} $ $ =\underset{h\to 0}{\mathop{\lim }},\frac{(1/h)\sqrt{4,-5h+8h^{2}}}{(1/h),(-,4+5h)}=\frac{\sqrt{4}}{-4}=-\frac{1}{2} $ .
BETA
