If 0< x< y, then _ n (y^n+x^n )^ 1 n = - Vidyadip

MCQ

If $0< x< y$, then $\lim _{n \rightarrow \infty}\left(y^n+x^n\right)^{\frac{1}{n}}=$

A
1
B
$x$
✓
$y$
D
e

✓

Answer

Correct option: C.

$y$

(C)
Since $0< x< y$
$\therefore \quad 0<\frac{x}{y}<1$
$\therefore \quad \lim _{ n \rightarrow \infty}\left(\frac{x}{y}\right)^{ n }=0$
$\therefore \quad \lim _{n \rightarrow \infty}\left(y^n+x^n\right)^{\frac{1}{n}}=\lim _{n \rightarrow \infty} y\left[1+\left(\frac{x}{y}\right)^n\right]^{\frac{1}{n}}$
$=y(1+0)^0=y$

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