Question

If $0<x<y,$ then ${lim}_{n\to \infty }{(y^n+x^n)}^{1/n}$ is equal to :

• A. $e$
• B. $x$
• C. $y$
• D. None of these

Answer 1

The correct option is C $y$

Given $x,y>0$ it is max $\{x,y\}$

This is obvious for x=y lines then max $\{x,y\}$ = x = y

Also it is obvious if either is 0.

$0<x<y$

$(y^n+x^n{)}^{1/n}$

$y(1+(\frac{x}{y}{)}^n{)}^{1/n}$

$\frac{x}{y}$ = P.

$y(1+P^n{)}^{1/n}$

take a fixed s such that $0<s<p.$

for certain value N, $p^n\le s$ for all $n\ge N$ which

means that $1<(1+p^n{)}^{1/n}\le (1+s{)}^{1/n}forn\ge N$

& we know that

${lim}_{n\to \infty }(1+s{)}^{1/n}=1.$

Become${lim}_{n\to \infty }\frac{1}{n}(1+s)=0$

Hence by using square therm

${lim}_{n\to \infty }(y^n+x^n{)}^{1/n}=y{lim}_{n\to \infty }(1+s{)}^{1/n}=y$

Ans :- y