Functions Question 166
Question: If $ 0<x<y $ then $ \underset{n\to \infty }{\mathop{\lim }},{{(y^{n}+x^{n})}^{1/n}} $ is equal to
Options:
A) $ e $
B) $ x $
C) $ y $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
We have $ \underset{n\to \infty }{\mathop{\lim }}{{(x^{n}+y^{n})}^{1/n}}=y\underset{n\to \infty }{\mathop{\lim }}{{( 1+{{( \frac{x}{y} )}^{n}} )}^{1/n}} $ $ =y\underset{n\to \infty }{\mathop{\lim }},{{[ 1+{{( \frac{x}{y} )}^{n}} ]}^{{{( \frac{y}{x} )}^{n}}.\frac{1}{n}.{{( \frac{x}{y} )}^{n}}}} $ $ =y\underset{n\to \infty }{\mathop{\lim }},{{[ {{( 1+{{( \frac{x}{y} )}^{n}} )}^{{{( \frac{y}{x} )}^{n}}.}} ]}^{\frac{1}{n}.{{( \frac{x}{y} )}^{n}}}} $ $ =ye^{0}=y $ , $ [ \because \frac{x}{y}<1,\Rightarrow {{( \frac{x}{y} )}^{n}}\to 0asn\to \infty ] $ .
BETA
