Differentiation Question 152
Question: The value of $ \underset{x\to 0}{\mathop{\lim }}\frac{27^{x}-9^{x}-3^{x}+1}{\sqrt{2}-\sqrt{1+\cos x}} $ is
Options:
A) $ 4\sqrt{2}{{(log3)}^{2}} $
B) $ 8\sqrt{2}{{(log3)}^{2}} $
C) $ 2\sqrt{2}{{(log3)}^{2}} $
D) None of these
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Answer:
Correct Answer: B
Solution:
[b] $ \underset{x\to 0}{\mathop{\lim }}\frac{27^{x}-9^{x}-3^{x}+1}{\sqrt{2}-\sqrt{1+\cos x}} $
$ =\underset{x\to 0}{\mathop{\lim }}\frac{9^{x}{{.3}^{x}}-9^{x}-3^{x}+1}{\sqrt{2}-\sqrt{2}\cos \frac{x}{2}} $
$ =\underset{x\to 0}{\mathop{\lim }}( \frac{9^{x}-1}{x} ).( \frac{3^{x}-1}{x} ).\frac{1}{\sqrt{2}}.x^{2}.\frac{1}{2{{\sin }^{2}}\frac{x}{4}} $
$ =\underset{x\to 0}{\mathop{\lim }}( \frac{9^{x}-1}{x} ).( \frac{3^{x}-1}{x} ).\frac{1}{\sqrt{2}}( \frac{x^{2}/16}{{{\sin }^{2}}x/4} )8 $
$ =\frac{8}{\sqrt{2}}(log9)(log3)=8\sqrt{2}{{(log3)}^{2}}. $
BETA
