Functions Question 264
Question: $ \underset{x\to 0}{\mathop{\lim }},\frac{4^{x}-9^{x}}{x(4^{x}+9^{x})}= $
[EAMCET 2002]
Options:
A) $ \log ( \frac{2}{3} ) $
B) $ \frac{1}{2}\log ( \frac{3}{2} ) $
C) $ \frac{1}{2}\log ( \frac{2}{3} ) $
D) $ \log ,( \frac{3}{2} ) $
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Answer:
Correct Answer: A
Solution:
$ y=\underset{x\to 0}{\mathop{\lim }},\frac{4^{x}-9^{x}}{x(4^{x}+9^{x})} $ , $ ( \frac{0}{0}form ) $ Using L-Hospital?s rule, $ y=\underset{x\to 0}{\mathop{\lim }},\frac{4^{x}\log 4-9^{x}\log 9}{(4^{x}+9^{x})+x(4^{x}\log 4+9^{x}\log 9)} $
Þ $ y=\frac{\log 4-\log 9}{2} $
Þ $ y=\frac{\log {{( \frac{2}{3} )}^{2}}}{2}=\log \frac{2}{3} $ .
BETA
