Differentiation Question 473
Question: The value of $ \underset{x\to 0}{\mathop{\lim }},\frac{{{(4^{x}-1)}^{3}}}{\sin \frac{x^{2}}{4}\log (1+3x)}, $ is
Options:
A) $ \frac{4}{3}{{(in4)}^{2}} $
B) $ \frac{4}{3}{{(In4)}^{3}} $
C) $ \frac{3}{2}{{(In4)}^{2}} $
D) $ \frac{3}{2}{{(In4)}^{3}} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ \underset{x\to 0}{\mathop{\lim }},\frac{{{(4^{x}-1)}^{3}}}{\sin \frac{x^{2}}{4}\log (1+3x)} $ $ =\underset{x\to 0}{\mathop{\lim }},\frac{{{(4^{x}-1)}^{3}}}{x^{3}}.\frac{{{(x/2)}^{2}}}{\sin x^{2}/4}.\frac{3x}{\log (1+3x)}.\frac{4}{3} $ $ =\frac{4}{3}{{(log_{e}4)}^{3}}.1.{\log_{e}}(e)=\frac{4}{3}{{(log_{e}4)}^{3}}. $
BETA
