Functions Question 295
Question: The value of $ \underset{x\to \infty }{\mathop{\lim }},\frac{(x+1)(3x+4)}{x^{2}(x-8)} $ is equal to
[Pb. CET 2002]
Options:
A) 2
B) 3
C) 1
D) 0
Show Answer
Answer:
Correct Answer: D
Solution:
$ \underset{x\to \infty }{\mathop{\lim }},\frac{(x+1)(3x+4)}{x^{2}(x-8)}=\underset{x\to \infty }{\mathop{\lim }},[ \frac{x( 1+\frac{1}{x} ),x,( 3+\frac{4}{x} )}{x^{3}( 1-\frac{8}{x} )} ] $ $ =\underset{x\to \infty }{\mathop{\lim }},[ \frac{1}{x}\frac{( 1+\frac{1}{x} ),( 3+\frac{4}{x} )}{( 1-\frac{8}{x} )} ]=0 $ .
BETA
