Functions Question 266
Question: $ \underset{x\to 0}{\mathop{\lim }},\frac{a^{x}-b^{x}}{e^{x}-1} $ =
[Kerala (Engg.) 2002]
Options:
A) $ \log ( \frac{a}{b} ) $
B) $ \log ( \frac{b}{a} ) $
C) $ \log (a,b) $
D) $ \log ,(a+,b) $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \underset{x\to 0}{\mathop{\lim }},\frac{a^{x}-b^{x}}{e^{x}-1}=\underset{x\to 0}{\mathop{\lim }},\frac{a^{x}-b^{x}}{x}.\frac{x}{e^{x}-1} $ $ =\underset{x\to 0}{\mathop{\lim }},[ \frac{a^{x}-1}{x}-\frac{b^{x}-1}{x} ]\frac{x}{e^{x}-1} $ $ =({\log_{e}}a-{\log_{e}}b).\frac{1}{{\log_{e}}e} $ $ ={\log_{e}}( \frac{a}{b} ) $ Trick : Apply L-Hospital?s rule.
BETA
