MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Hess' Law

If 0<x<y th...

Question

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

Open in App

Solution

$\begin{matrix} {(x^{n} + y^{n})}^{\frac{1}{n}} = y {(1 + \in^{n})}^{\frac{1}{h}} \Rightarrow t a k e f i x e d i \cdot i < δ \leq E [E = \frac{x}{y}] \Rightarrow E^{_{n}} \leq δ, f o r a l l n \geq N \Rightarrow 1 < {(1 + E^{n})}^{\frac{1}{n}} \leq {(1 + δ)}^{\frac{1}{n}} f o r n \geq N \Rightarrow {lim}_{x \to \infty} {(1 + δ)}^{\frac{1}{n}} = 1, {lim}_{x \to \infty} \frac{1}{n} ln (1 + δ) = 0 \Rightarrow {lim}_{x \to \infty} {(x^{n} + y^{h})}^{\frac{1}{n}} = y {lim}_{x \to \infty} (1 + δ) = y \end{matrix}$

Suggest Corrections

thumbs-up

0

Similar questions

0 < x < y

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

0 < x < y,

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

y (n) = e^{x} e^{x^{2}} . . . e^{x^{n}}, 0 < x < 1

lim n \to \infty \frac{d y (n)}{d x}

x = \frac{1}{2}

y (n) = e^{x} e^{x^{2}} . . . e^{x^{n}}, 0 < x < 1

lim n \to \infty \frac{d y (n)}{d x}

\frac{1}{2}

a = \sum_{n = 0}^{\infty} x^{n}, b = \sum_{n = 0}^{\infty} y^{n}, c = \sum_{n = 0}^{\infty} (x y)^{n}

| x |, | y | < 1

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Thermochemistry

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon