Functions Question 604
Question: If $ \underset{x\to a}{\mathop{\lim }},\frac{a^{x}-x^{a}}{x^{x}-a^{a}}=-1 $ , then
[EAMCET 2003]
Options:
A) $ a=1 $
B) $ a=0 $
C) $ a=e $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Using L-Hospital’s rule, we get
$ -1=\underset{x\to a}{\mathop{\lim }},\frac{a^{x}-x^{a}}{x^{x}-a^{a}}=\underset{x\to a}{\mathop{\lim }}\frac{a^{x}{\log_{e}}a-a{x^{a-1}}}{x^{x}+a^{a}{\log_{e}}a} $
$ \Rightarrow -1=\frac{a^{a}{\log_{e}}a-a,.,{a^{a-1}}}{a^{a}+a^{a}{\log_{e}}a}=\frac{{\log_{e}}a-1}{{\log_{e}}a+1} $ …..(i)
Now (i) is satisfied only when $ a=1. $
BETA
