Differentiation Question 155
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}} $
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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
