Question

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

Solution

## Let $$x = \dfrac{1}{y}$$ so that $$y > 1$$Also, let $$y^{n} = z$$$$\Rightarrow n\log y = \log z$$Then,$$nx^{n} = \dfrac{n}{y^{n}} =\dfrac{1}{z}.\dfrac{\log z}{\log y} =\dfrac{1}{\log y} .\dfrac{\log z}{z}$$Now, when $$n$$ is infinite, $$z$$ is infinite and$$\dfrac{\log z}{z}=0$$Also, $$\log y$$ is finiteTherefore, $$\displaystyle \lim_{n \rightarrow \infty} nx^{n} =0$$Mathematics

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