Question

The value of $\mathop {\lim }\limits_{n \to \infty } {\left( {{6^n} + {5^n}} \right)^{\dfrac{1}{n}}}$ is equal to __
A). 6
B). 5
C). $\dfrac{5}{6}$
D). e

Answer 1

Hint: Here, to find the value of $\mathop {\lim }\limits_{n \to \infty } {\left( {{6^n} + {5^n}} \right)^{\dfrac{1}{n}}}$, we will first take out ${5^n}$ from the bracket by dividing both the terms with ${5^n}$. Then we will take the power of $\dfrac{{{6^n}}}{{{5^n}}}$ ass common. Now, since $\dfrac{6}{5}$ is greater than 1 and n tends to infinity, ${\left( {\dfrac{6}{5}} \right)^n}$ will be very much larger than 1 and so we can ignore 1. Now, just simplify and we will get our answer.

Complete step-by-step solution:
In this question, we are given a limit and we need to find its value.
Given: $\mathop {\lim }\limits_{n \to \infty } {\left( {{6^n} + {5^n}} \right)^{\dfrac{1}{n}}}$- - - - - - - - - - - - - (1)
Here, we are asked that what will be the value of ${\left( {{6^n} + {5^n}} \right)^{\dfrac{1}{n}}}$ when n tends to infinity.
Let us solve this question.
Here, in equation (1), we can take out ${5^n}$ from the bracket by dividing the both terms with ${5^n}$. Therefore, we get
$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } {\left( {{6^n} + {5^n}} \right)^{\dfrac{1}{n}}} = \mathop {\lim }\limits_{n \to \infty } {\left( {{5^n}\left( {\dfrac{{{6^n}}}{{{5^n}}} + 1} \right)} \right)^{\dfrac{1}{n}}} = \mathop {\lim }\limits_{n \to \infty } {\left( {{5^n}} \right)^{\dfrac{1}{n}}}{\left( {\dfrac{{{6^n}}}{{{5^n}}} + 1} \right)^{\dfrac{1}{n}}}$
Now, n gets cancelled and we can take n power common in $\dfrac{{{6^n}}}{{{5^n}}}$. Therefore, we get
$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } {\left( {{6^n} + {5^n}} \right)^{\dfrac{1}{n}}} = \mathop {\lim }\limits_{n \to \infty } \left( 5 \right){\left( {{{\left( {\dfrac{6}{5}} \right)}^n} + 1} \right)^{\dfrac{1}{n}}}$
Now, as n tends to infinity and the value of $\dfrac{6}{5}$ is greater than 1, the value of ${\left( {\dfrac{6}{5}} \right)^n}$ will be very much larger than 1. In fact it will be such big, that we can neglect 1. So, therefore, we get
$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } {\left( {{6^n} + {5^n}} \right)^{\dfrac{1}{n}}} = \mathop {\lim }\limits_{n \to \infty } \left( 5 \right){\left( {{{\left( {\dfrac{6}{5}} \right)}^n}} \right)^{\dfrac{1}{n}}}$
Here, again n gets cancelled. Therefore, we get
$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } {\left( {{6^n} + {5^n}} \right)^{\dfrac{1}{n}}} = \left( 5 \right)\left( {\dfrac{6}{5}} \right) = 6$
Therefore, the value of $\mathop {\lim }\limits_{n \to \infty } {\left( {{6^n} + {5^n}} \right)^{\dfrac{1}{n}}}$ is equal to 6.
Hence, option A is the correct answer.

Note: Note that we cannot directly put the value of n as infinity. If we do so, we won’t get our answer. Let us see what we get if we put the value of n before solving the expression.
$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } {\left( {{6^n} + {5^n}} \right)^{\dfrac{1}{n}}} = {\left( {{6^\infty } + {5^\infty }} \right)^{\dfrac{1}{\infty }}} = {\left( {\infty + \infty } \right)^0}$
Here, we get an answer in the form ${\infty ^0}$. So, to avoid this form, we need to simplify our expression first.