Differentiation Question 486
Question: The value of $ \underset{x\to \infty }{\mathop{\lim }},\frac{({2^{x^{n}}}){e^{\frac{1}{^{x}}}}-({3^{x^{n}}}){e^{\frac{1}{x}}}}{x^{n}} $ (where $ n\in N $ ) is
Options:
A) $ \log n( \frac{2}{3} ) $
B) 0
C) $ n\log n( \frac{2}{3} ) $
D) Not defined
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ L=\underset{x\to \infty }{\mathop{\lim }},\frac{{{({2^{x^{n}}})}^{\frac{1}{e^{x}}}}-{{({3^{x^{n}}})}^{\frac{1}{e^{x}}}}}{x^{n}}=\underset{x\to \infty }{\mathop{\lim }},\frac{(3){{,}^{^{\frac{x^{n}}{e^{x}}}}}( {{( \frac{2}{3} )}^{\frac{x^{n}}{e^{x}}}}-1 )}{x^{n}} $ Now, $ \underset{x\to \infty }{\mathop{\lim }},\frac{x^{n}}{e^{x}}=\underset{x\to \infty }{\mathop{\lim }},\frac{n!}{e^{x}}=0 $ (Applying L?Hospital?s rule n times) Hence, $ L=\underset{x\to \infty }{\mathop{\lim }},{{(3)}^{\frac{x^{n}}{e^{x}}}}\underset{x\to \infty }{\mathop{\lim }},\frac{( {{( \frac{2}{3} )}^{\frac{x^{n}}{e^{x}}}}-1 )}{\frac{x^{n}}{e^{x}}}\underset{x\to \infty }{\mathop{\lim }},\frac{1}{e^{x}} $ $ =1\times \log (2/3)\times 0=0. $
BETA
