Functions Question 181
Question: The value of $ \underset{x\to {0^{+}}}{\mathop{\lim }},x^{m}{{(\log x)}^{n}},\ m,\ n\in N $ is
Options:
A) 0
B) $ \frac{m}{n} $
C) $ mn $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \underset{x\to 0+}{\mathop{\lim }}x^{m},{{(\log x)}^{n}}=\underset{x\to 0+}{\mathop{\lim }}\frac{{{(\log x)}^{n}}}{{x^{-m}}} $ $ ( Form\frac{\infty }{\infty } ) $ $ \log P=\int_0^{1}{{}}\log x,dx=(x,\log x-x)_0^{1}=(-1) $ (By L-Hospital’s rule) $ =\underset{x\to 0+}{\mathop{\lim }}\frac{n,{{(\log x)}^{n-1}}}{-m{x^{-m}}}, $ $ ( Form\frac{\infty }{\infty } ) $ $ =\underset{x\to 0+}{\mathop{\lim }}\frac{n,(n-1),{{(\log x)}^{(n-2)}}\frac{1}{x}}{{{(-m)}^{2}}{x^{-m-1}}} $ (By L-Hospital’s rule) $ =\underset{x\to 0+}{\mathop{\lim }}\frac{n,(n-1),{{(\log x)}^{n-2}}}{m^{2}{x^{-m}}}, $ $ ( Form\frac{\infty }{\infty } ) $ ………………….. …………………. $ =\underset{x\to 0+}{\mathop{\lim }}\frac{n!}{{{(-m)}^{n}}{x^{-m}}}=0 $ (Differentiating $ N^{r} $ and $ D^{r} $ n times).
BETA
