Functions Question 63
Question: $ \underset{x\to \infty }{\mathop{\lim }},[x({a^{1/x}}-1)] $ , $ (a>1)= $
Options:
A) $ \log x $
B) 1
C) 0
D) $ -\log \frac{1}{a} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \underset{x\to \infty }{\mathop{\lim }},x,({a^{1/x}}-1)=\underset{x\to \infty }{\mathop{\lim }}[ \frac{{a^{1/x}}-1}{1/x} ] $ $ =\underset{x\to \infty }{\mathop{\lim }},\frac{[{e^{{\log_{e}}{a^{1/x}}}}-1]}{1/x}={\log_{e}}a=-{\log_{e}}\frac{1}{a}. $
BETA
