Differentiation Question 117
Question: If $ x>0 $ and $ g $ is a bounded function, then $ \underset{n\to \infty }{\mathop{\lim }}\frac{f(x)e^{nx}+g(x)}{e^{nx}+1} $ is
Options:
A) 0
B) $ f(x) $
C) $ g(x) $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Given that, x > 0 and g(x) is bounded function. $ Limit=\underset{n\to \infty }{\mathop{\lim }}\frac{f(x)e^{nx}+g(x)}{e^{nx}+1} $
$ =\underset{n\to \infty }{\mathop{\lim }}\frac{f(x)}{1+( \frac{1}{e^{nx}} )}+\frac{g(x)}{e^{nx}+1}=\frac{f(x)}{1+0}+\frac{finite}{\infty } $
$ =f(x) $
BETA
