What is the limit as k approaches infinity of (((k+1)^{75}(k+4)))/((k+5)k^{75)} ?

The limit as k approaches infinity of (((k+1)^{75}(k+4)))/((k+5)k^{75)} is 1

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limit as k approaches infinity of (((k+1)^{75}(k+4)))/((k+5)k^{75)}

k \to \infty lim \frac{( ( k + 1 ) ^{75} ( k + 4 ) )}{( k + 5 ) k ^{75}}

1

Solution steps

k \to \infty lim (\frac{( k + 1 ) ^{75} ( k + 4 )}{( k + 5 ) k ^{75}})

Apply the following algebraic property

: a + b = a (1 + \frac{b}{a})

\frac{( k + 1 ) ^{75} ( k + 4 )}{( k + 5 ) k ^{75}} = \frac{( k ( 1 + \frac{1}{k} ) ) ^{75} ( k ( 1 + \frac{4}{k} ) )}{( k ( 1 + \frac{5}{k} ) ) k ^{75}}

= k \to \infty lim (\frac{( k ( 1 + \frac{1}{k} ) ) ^{75} k ( 1 + \frac{4}{k} )}{k ( 1 + \frac{5}{k} ) k ^{75}})

Simplify

\frac{( k ( 1 + \frac{1}{k} ) ) ^{75} ( k ( 1 + \frac{4}{k} ) )}{( k ( 1 + \frac{5}{k} ) ) k ^{75}} : \frac{( \frac{1}{k} + 1 ) ^{75} ( \frac{4}{k} + 1 )}{\frac{5}{k} + 1}

= k \to \infty lim (\frac{( \frac{1}{k} + 1 ) ^{75} ( \frac{4}{k} + 1 )}{\frac{5}{k} + 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{lim _{k \to \infty} ( ( \frac{1}{k} + 1 ) ^{75} ( \frac{4}{k} + 1 ) )}{lim _{k \to \infty} ( \frac{5}{k} + 1 )}

k \to \infty lim ((\frac{1}{k} + 1)^{75} (\frac{4}{k} + 1)) = 1

k \to \infty lim (\frac{5}{k} + 1) = 1

= \frac{1}{1}

Simplify

= 1

x \to \infty lim (ln (3))

x^{\frac{1}{3}}

x \to 4 lim (5 - (- 7))

y = 4 e^{x}

What is the limit as k approaches infinity of (((k+1)^{75}(k+4)))/((k+5)k^{75)} ?
The limit as k approaches infinity of (((k+1)^{75}(k+4)))/((k+5)k^{75)} is 1