What is the limit as n approaches infinity of (n^{sqrt(2)})/(21^n) ?

The limit as n approaches infinity of (n^{sqrt(2)})/(21^n) is 0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

limit as n approaches infinity of (n^{sqrt(2)})/(21^n)

n \to \infty lim (\frac{n ^{2}}{2 1 ^{n}})

0

Solution steps

n \to \infty lim (\frac{n ^{2}}{2 1 ^{n}})

Apply L'Hopital's Rule

: n \to \infty lim (\frac{2 n ^{2 - 1}}{2 1 ^{n} ln ( 21 )})

= n \to \infty lim (\frac{2 n ^{2 - 1}}{2 1 ^{n} ln ( 21 )})

Apply L'Hopital's Rule

: n \to \infty lim (\frac{2 ( 2 - 1 ) n ^{2 - 2}}{ln ^{2} ( 21 ) \cdot 2 1 ^{n}})

= n \to \infty lim (\frac{2 ( 2 - 1 ) n ^{2 - 2}}{ln ^{2} ( 21 ) \cdot 2 1 ^{n}})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= \frac{2 ( 2 - 1 )}{ln ^{2} ( 21 )} \cdot n \to \infty lim (\frac{n ^{2 - 2}}{2 1 ^{n}})

Expand

2 (2 - 1) : 2 - 2

= \frac{2 - 2}{ln ^{2} ( 21 )} \cdot n \to \infty lim (\frac{n ^{2 - 2}}{2 1 ^{n}})

x \to a lim [\frac{f ( x )}{g ( x )}] = \frac{lim _{x \to a} f ( x )}{lim _{x \to a} g ( x )}, x \to a lim g (x) \neq = 0

With the exception of indeterminate form

= \frac{2 - 2}{ln ^{2} ( 21 )} \cdot \frac{lim _{n \to \infty} ( n ^{2 - 2} )}{lim _{n \to \infty} ( 2 1 ^{n} )}

n \to \infty lim (n^{2 - 2}) = 0

n \to \infty lim (2 1^{n}) = \infty

= \frac{2 - 2}{ln ^{2} ( 21 )} \cdot \frac{0}{\infty}

Simplify

\frac{2 - 2}{ln ^{2} ( 21 )} \cdot \frac{0}{\infty} : 0

= 0

5 t^{2} - 2 e^{t}

\frac{\partial}{\partial y} (4 x^{3} - 4 y)

\frac{d}{d x} (- \frac{k}{x ^{2}})

\int \frac{2 x + 3}{x + 1} d x

What is the limit as n approaches infinity of (n^{sqrt(2)})/(21^n) ?
The limit as n approaches infinity of (n^{sqrt(2)})/(21^n) is 0