What is the limit as n approaches infinity of (((n+1)^3))/(5^{n+1)} ?

The limit as n approaches infinity of (((n+1)^3))/(5^{n+1)} is 0

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limit as n approaches infinity of (((n+1)^3))/(5^{n+1)}

n \to \infty lim \frac{( ( n + 1 ) ^{3} )}{5 ^{n + 1}}

0

Solution steps

n \to \infty lim (\frac{( n + 1 ) ^{3}}{5 ^{n + 1}})

Apply L'Hopital's Rule

: n \to \infty lim (\frac{3 ( n + 1 ) ^{2}}{ln ( 5 ) \cdot 5 ^{n + 1}})

= n \to \infty lim (\frac{3 ( n + 1 ) ^{2}}{ln ( 5 ) \cdot 5 ^{n + 1}})

Apply L'Hopital's Rule

: n \to \infty lim (\frac{6 ( n + 1 )}{ln ^{2} ( 5 ) \cdot 5 ^{n + 1}})

= n \to \infty lim (\frac{6 ( n + 1 )}{ln ^{2} ( 5 ) \cdot 5 ^{n + 1}})

Apply L'Hopital's Rule

: n \to \infty lim (\frac{6}{ln ^{3} ( 5 ) \cdot 5 ^{n + 1}})

= n \to \infty lim (\frac{6}{ln ^{3} ( 5 ) \cdot 5 ^{n + 1}})

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

= \frac{6}{ln ^{3} ( 5 )} \cdot n \to \infty lim (\frac{1}{5 ^{n + 1}})

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{6}{ln ^{3} ( 5 )} \cdot \frac{lim _{n \to \infty} ( 1 )}{lim _{n \to \infty} ( 5 ^{n + 1} )}

n \to \infty lim (1) = 1

n \to \infty lim (5^{n + 1}) = \infty

= \frac{6}{ln ^{3} ( 5 )} \cdot \frac{1}{\infty}

Simplify

\frac{6}{ln ^{3} ( 5 )} \cdot \frac{1}{\infty} : 0

= 0

x^{2} y^{^{''}} + 5 x y^{^{'}} + 3 y = 0

\frac{d}{d x} (x^{8 x})

x \to 0 lim (- arctan (\frac{x}{2}))

f^{^{'}} (x) = - 4 x^{3} + 2 x^{2} - 7

f (x) = x + 2 cos (x), at x = 0

What is the limit as n approaches infinity of (((n+1)^3))/(5^{n+1)} ?
The limit as n approaches infinity of (((n+1)^3))/(5^{n+1)} is 0