Differentiation Question 504
Question: What is $ \underset{x\to 0}{\mathop{\lim }},\frac{\sin 2x+4x}{2x+\sin 4x} $ equal to?
Options:
A) 0
B) $ \frac{1}{2} $
C) 1
D) 2
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ \underset{x\to 0}{\mathop{\lim }},\frac{\sin 2x+4x}{2x+\sin 4x}=\underset{x\to 0}{\mathop{\lim }},\frac{\frac{\sin 2x}{x}+4}{2+\frac{\sin 4x}{x}} $ $ =\underset{x\to 0}{\mathop{\lim }},\frac{2( \frac{\sin 2x}{2x} )+4}{2+4( \frac{\sin 4x}{4x} )} $ Applying limit, we get $ \frac{2+4}{2+4}=1 $
BETA
