# I'm wondering about the motivation for proving the limit. <mstyle displaystyle="true" scriptleve

I'm wondering about the motivation for proving the limit.
$\underset{x\to 1}{lim}\sqrt[n]{x}=1$
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xxsailojaixxv5
Lets go to the definition of the limit.
$\mathrm{\forall }ϵ>0,\mathrm{\exists }\delta >0:|x-1|<\delta \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}|\sqrt[n]{x}-1|<ϵ$
The "trick" here will be very similar to the one above, but it might look more natural here.
Multiplying by "1."
$|\frac{\sqrt[n]{x}-1}{1}\frac{1+\sqrt[n]{x}+{\sqrt[n]{x}}^{2}+\cdots +{\sqrt[n]{x}}^{n-1}}{1+\sqrt[n]{x}+{\sqrt[n]{x}}^{2}+\cdots +{\sqrt[n]{x}}^{n-1}}|=|\frac{x-1}{1+\sqrt[n]{x}+{\sqrt[n]{x}}^{2}+\cdots +{\sqrt[n]{x}}^{n-1}}|$
let $\delta \le 1$ then $x>0$ and $1+\sqrt[n]{x}+{\sqrt[n]{x}}^{2}+\cdots +{\sqrt[n]{x}}^{n-1}>1$
$|\sqrt[n]{x}-1|<|x-1|<\delta$