Show that when n is infinite the limit of $n{x^n}$ tends to 0, when x>1.

147.6k+ views
Hint: To solve this problem we need to have knowledge about the limits concept and its value substitution. Here we have taken x value where y should be greater than one.

Complete step-by-step answer:
Here let us consider the value x as
$ \Rightarrow x = \dfrac{1}{y}$
So here we can that $y > 1$

Now let us take ${y^n} = z$
$ \Rightarrow {y^n} = z$

On applying log on both sides we get
$ \Rightarrow \log {y^n} = \log z$
$ \Rightarrow n\log y = \log z$

$n{x^n} = \dfrac{n}{{{y^n}}} = \dfrac{1}{z}.\dfrac{{\log z}}{{\log y}} = \dfrac{1}{{\log y}}.\dfrac{{\log z}}{z}$

Now here when n is infinite then z is also infinite.

Here by using above condition we can say that
$ \Rightarrow \dfrac{{\log z}}{z} = 0$

Also from this we can say that log y is infinite.

Therefore, $\mathop {\lim }\limits_{x \to \infty } n{x^n} = 0$.

Note: The above question is based on limits based. Here we have taken the value of x where y value should be greater than 1, later we took one more term ${y^n} = z$ using this term we have shown that
$\mathop {\lim }\limits_{x \to \infty } n{x^n} = 0$.