Question

Find the value of \[\mathop {\lim }\limits_{x \to \infty } {\left( {\dfrac{{{1^{1/x}} + {2^{1/x}} + {3^{1/x}} + .... + {n^{1/x}}}}{n}} \right)^{nx}}\] , $n \in \mathbb{N}$ is equal to
A. $n!$
B.1
C. $\dfrac{1}{{n!}}$
D.0

Answer 1

Hint: We check the value of the limit by applying the limit directly. As the limit tends to ${1^\infty }$ ,
We can apply the relation that $\mathop {\lim }\limits_{x \to a} f{\left( x \right)^{g\left( x \right)}} = {e^{\mathop {\lim }\limits_{x \to a} g\left( x \right)\left[ {f\left( x \right) - 1} \right]}}$ . Then we can find the power of e. By directly applying the limits we get $\dfrac{0}{0}$ , so we can apply L hospital's rule which is given by $\mathop {\lim }\limits_{x \to \infty } \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$ . Then by applying the properties of the logarithms we get the required limit.

Complete step-by-step answer:
We need to find the limit of $\mathop {\lim }\limits_{x \to \infty } {\left( {\dfrac{{{1^{1/x}} + {2^{1/x}} + {3^{1/x}} + .... + {n^{1/x}}}}{n}} \right)^{nx}}$
We can check what the values of the limit at infinity by substituting for infinity at all the x.
 $ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {\left( {\dfrac{{{1^{1/x}} + {2^{1/x}} + {3^{1/x}} + .... + {n^{1/x}}}}{n}} \right)^{nx}} = {\left( {\dfrac{{{1^{1/\infty }} + {2^{1/\infty }} + {3^{1/\infty }} + .... + {n^{1/\infty }}}}{n}} \right)^{n\infty }}$
We know that $\dfrac{1}{\infty } = 0$ ,
 $ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {\left( {\dfrac{{{1^{1/x}} + {2^{1/x}} + {3^{1/x}} + .... + {n^{1/x}}}}{n}} \right)^{nx}} = {\left( {\dfrac{{{1^0} + {2^0} + {3^0} + .... + {n^0}}}{n}} \right)^{n\infty }}$
We know that ${a^0} = 1$ . So, we get,
 \[ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {\left( {\dfrac{{{1^{1/x}} + {2^{1/x}} + {3^{1/x}} + .... + {n^{1/x}}}}{n}} \right)^{nx}} = {\left( {\dfrac{{1 + 1 + 1 + .... + 1}}{n}} \right)^{n\infty }}\]
There are n terms in the numerator, so the limit will become,
 $ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {\left( {\dfrac{{{1^{1/x}} + {2^{1/x}} + {3^{1/x}} + .... + {n^{1/x}}}}{n}} \right)^{nx}} = {\left( {\dfrac{n}{n}} \right)^{n\infty }}$
On simplification we get,
 $ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {\left( {\dfrac{{{1^{1/x}} + {2^{1/x}} + {3^{1/x}} + .... + {n^{1/x}}}}{n}} \right)^{nx}} = {1^\infty }$
Now the limit is ${1^\infty }$ .
We know that for limits that tends to ${1^\infty }$ ,
If $\mathop {\lim }\limits_{x \to a} f{\left( x \right)^{g\left( x \right)}} = {1^\infty }$ , then its limit is given by the equation, ${e^{\mathop {\lim }\limits_{x \to a} g\left( x \right)\left[ {f\left( x \right) - 1} \right]}}$
We can use this in the given function.
Let $L = \mathop {\lim }\limits_{x \to \infty } {\left( {\dfrac{{{1^{1/x}} + {2^{1/x}} + {3^{1/x}} + .... + {n^{1/x}}}}{n}} \right)^{nx}}$
Here $f\left( x \right) = \dfrac{{{1^{1/x}} + {2^{1/x}} + {3^{1/x}} + .... + {n^{1/x}}}}{n}$ and $g\left( x \right) = nx$ . By using the above property, we get,
 $ \Rightarrow L = {e^{\mathop {\lim }\limits_{x \to \infty } nx\left( {\dfrac{{{1^{1/x}} + {2^{1/x}} + {3^{1/x}} + .... + {n^{1/x}}}}{n} - 1} \right)}}$
Let us take \[ \Rightarrow I = \mathop {\lim }\limits_{x \to \infty } nx\left( {\dfrac{{{1^{1/x}} + {2^{1/x}} + {3^{1/x}} + .... + {n^{1/x}}}}{n} - 1} \right)\]
 $ \Rightarrow L = {e^I}$ …………….… (1)
We can cancel the n from both numerator and denominator,
 $ \Rightarrow I = \mathop {\lim }\limits_{x \to \infty } x\left( {{1^{1/x}} + {2^{1/x}} + {3^{1/x}} + .... + {n^{1/x}} - n} \right)$
We can write x in the numerator as $\dfrac{1}{x}$ in the denominator.
 $ \Rightarrow I = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{{1^{1/x}} + {2^{1/x}} + {3^{1/x}} + .... + {n^{1/x}} - n}}{{1/x}}} \right)$
On applying the limits, we get,
 $ \Rightarrow I = \left( {\dfrac{{{1^{1/\infty }} + {2^{1/\infty }} + {3^{1/\infty }} + .... + {n^{1/\infty }} - n}}{{1/\infty }}} \right)$
We know that $\dfrac{1}{\infty } = 0$ and ${a^0} = 1$ . Then the limit will become,
 $ \Rightarrow I = \left( {\dfrac{{1 + 1 + 1 + .... + 1 - n}}{0}} \right)$
There are n terms of 1 in the numerator, so the limit will become,
 $ \Rightarrow I = \left( {\dfrac{0}{0}} \right)$
As the limit is $\dfrac{0}{0}$ , we can apply L hospital’s rule.
According to L hospital’s rule,
 $\mathop {\lim }\limits_{x \to \infty } \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$
Here $f\left( x \right) = {1^{1/x}} + {2^{1/x}} + {3^{1/x}} + .... + {n^{1/x}} - n$
Then $f'\left( x \right) = \dfrac{d}{{dx}}\left( {{1^{1/x}} + {2^{1/x}} + {3^{1/x}} + .... + {n^{1/x}} - n} \right)$
We know that $\dfrac{d}{{dx}}{a^x} = {a^x}\log \left( x \right)$ and on applying chain rule, we get,
 \[ \Rightarrow f'\left( x \right) = \left( {{1^{1/x}}\log \left( 1 \right)\left( { - \dfrac{1}{{{x^2}}}} \right) + {2^{1/x}}\log \left( 2 \right)\left( { - \dfrac{1}{{{x^2}}}} \right) + {3^{1/x}}\log \left( 3 \right)\left( { - \dfrac{1}{{{x^2}}}} \right) + .... + {n^{1/x}}\log \left( n \right)\left( { - \dfrac{1}{{{x^2}}}} \right) - 0} \right)\]
We can take the common term outside,
 $ \Rightarrow f'\left( x \right) = \left( { - \dfrac{1}{{{x^2}}}} \right)\left( {{1^{1/x}}\log \left( 1 \right) + {2^{1/x}}\log \left( 2 \right) + {3^{1/x}}\log \left( 3 \right) + .... + {n^{1/x}}\log \left( n \right)} \right)$
Here $g\left( x \right) = \dfrac{1}{x}$
 $g'\left( x \right) = - \dfrac{1}{{{x^2}}}$
So, the limit will become,
 $ \Rightarrow I = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}} \right)$
 $ \Rightarrow I = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{\left( { - \dfrac{1}{{{x^2}}}} \right)\left( {{1^{1/x}}\log \left( 1 \right) + {2^{1/x}}\log \left( 2 \right) + {3^{1/x}}\log \left( 3 \right) + .... + {n^{1/x}}\log \left( n \right)} \right)}}{{ - \dfrac{1}{{{x^2}}}}}} \right)$
We can cancel the common terms,
 $ \Rightarrow I = \mathop {\lim }\limits_{x \to \infty } \left( {\left( {{1^{1/x}}\log \left( 1 \right) + {2^{1/x}}\log \left( 2 \right) + {3^{1/x}}\log \left( 3 \right) + .... + {n^{1/x}}\log \left( n \right)} \right)} \right)$
On applying the limits, we get,
 \[ \Rightarrow I = \left( {{1^{1/\infty }}\log \left( 1 \right) + {2^{1/\infty }}\log \left( 2 \right) + {3^{1/\infty }}\log \left( 3 \right) + .... + {n^{1/\infty }}\log \left( n \right)} \right)\]
We know that $\dfrac{1}{\infty } = 0$ and ${a^0} = 1$ . Then the limit will become
 $ \Rightarrow I = \left( {\log \left( 1 \right) + \log \left( 2 \right) + \log \left( 3 \right) + .... + \log \left( n \right)} \right)$
We know that $\log a + \log b = \log \left( {ab} \right)$ . So, the limit will become,
 $ \Rightarrow I = \log \left( {1 \times 2 \times 3 \times ..... \times n} \right)$
We know that $1 \times 2 \times 3 \times ..... \times n = n!$
On substituting in equation (1), we get,
 $ \Rightarrow L = {e^{\log \left( {n!} \right)}}$
We know that ${e^{\log a}} = a$
 $ \Rightarrow L = n!$
 $ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {\left( {\dfrac{{{1^{1/x}} + {2^{1/x}} + {3^{1/x}} + .... + {n^{1/x}}}}{n}} \right)^{nx}} = n!$

So, the correct answer is option A.

Note: When we get an expression to find the limit, we can directly apply the limits to check whether it is defined. Then we can use different methods to simplify the expression and then apply the limits to get the required value of limit. We must use the relation $\mathop {\lim }\limits_{x \to a} f{\left( x \right)^{g\left( x \right)}} = {e^{\mathop {\lim }\limits_{x \to a} g\left( x \right)\left[ {f\left( x \right) - 1} \right]}}$ only if $\mathop {\lim }\limits_{x \to a} f{\left( x \right)^{g\left( x \right)}} = {1^\infty }$ . Similarly, we must use only the L’ Hospital’s rule $\mathop {\lim }\limits_{x \to \infty } \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}$ , only when the limit tends to $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$