Differentiation Question 177
Question: The value of $ \underset{x\to \frac{\pi }{2}}{\mathop{\lim }}{{[ {1^{1/{{\cos }^{2}}x}}+{2^{1/{{\cos }^{2}}x}}+…+{n^{1/cos^{2}x}} ]}^{{{\cos }^{2x}}}} $ is
Options:
A) 0
B) n
C) $ \infty $
D) $ \frac{n(n+1)}{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ \underset{x\to \frac{\pi }{2}}{\mathop{\lim }}{{[ {1^{1/{{\cos }^{2}}x}}+{2^{1/{{\cos }^{2}}x}}+…+{n^{1/{{\cos }^{2}}x}} ]}^{{{\cos }^{2}}x}} $
$ =\underset{t\to \infty }{\mathop{\lim }}{{(1^{t}+2^{t}+…+n^{t})}^{1/t}} $
$ [ Onputting\frac{1}{{{\cos }^{2}}x}=t\ge 1 ] $
$ =\underset{t\to \infty }{\mathop{\lim }}{{(n^{t})}^{1/t}}{{[ {{( \frac{1}{n} )}^{t}}+{{( \frac{2}{n} )}^{t}}+…+{{( \frac{n}{n} )}^{t}} ]}^{1/t}} $
$ =n\underset{t\to \infty }{\mathop{\lim }}{{[ {{( \frac{1}{n} )}^{t}}+{{( \frac{2}{n} )}^{t}}+…+{{( \frac{n}{n} )}^{t}} ]}^{1/t}} $
$ =n{{(0+0+…+1)}^{0}}=n. $
BETA
