mathop {lim }limits_{n to infty } {({4^n} + { | vedclass

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(B) Let $L = \mathop {\lim }\limits_{n 	o \infty } {({4^n} + {5^n})^{1/n}}$.We can factor out the term with the largest base,which is $5^n$:$L = \mat

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) Let $L = \mathop {\lim }\limits_{n 	o \infty } {({4^n} + {5^n})^{1/n}}$.We can factor out the term with the largest base,which is $5^n$:$L = \mathop {\lim }\limits_{n 	o \infty } {\left[ {5^n \left( {{{\left( {\frac{4}{5}} ight)}^n} + 1} ight)} ight]^{1/n}}$$L = \mathop {\lim }\limits_{n 	o \infty } 5 \cdot {\left( {1 + {{\left( {\frac{4}{5}} ight)}^n}} ight)^{1/n}}$As $n 	o \infty$,the term ${\left( {\frac{4}{5}} ight)^n} 	o 0$.Therefore,$L = 5 \cdot {(1 + 0)^0} = 5 \cdot 1 = 5$.

$\mathop {\lim }\limits_{n \to \infty } {({4^n} + {5^n})^{1/n}}$ is equal to

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