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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 finite
Therefore, $$\displaystyle \lim_{n \rightarrow \infty} nx^{n}  =0$$

Mathematics

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