Question

prove limit n tends to infinity n^1/n =1​

Answer 1

Answer: limit n tends to infinity n^1/n =1​

Step-by-step explanation:

As the limit n tends to infinity $\frac{n^{1} }{n} =1$

So let $\frac{1}{n} =x$

When $n\rightarrow \infty$, x=0

$\xrightarrow[x\rightarrow \infty]{lim}(\frac{1}{n})^{\frac{1}{n}}$

$=\xrightarrow[x\rightarrow 0]{lim}(x)^{x}\\=e(ln \xrightarrow[x\rightarrow 0]{lim(x)^{x}})\\=e(\xrightarrow[x\rightarrow 0]{lim}xln x)\\=e(\xrightarrow[x\rightarrow 0]{lim}\frac{lnx}{\frac{1}{x}})\\=e(\xrightarrow[x\rightarrow 0]{lim}\frac{\frac{d}{dx}(ln x)}{\frac{d}{dx}(\frac{1}{x})})$

So now we have an indeterminate form $\frac{\infty }{\infty }$ , where L's hospital rule applies,

$= e (\xrightarrow[x\rightarrow 0]{lim}\frac{\frac{1}{x}}{-\frac{1}{x^{2}}})\\=e(\xrightarrow[x\rightarrow 0]{lim}(-x))\\=e^{0}\\=1$

Hence, limit n tends to infinity n^1/n =1​

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