Prove that $\lim_{x\to0^+}[1+[x]]^{\frac{2}{x}}=1$,where $[x]$ represents the floor function of $x$
$\lim_{x\to0^+}[1+[x]]^{\frac{2}{x}}=\lim_{x\to0^+}[1]^{\frac{2}{x}}$
Because $\lim_{x\to0^+}[x]=0$
But i am stuck.Please help me.Thanks.
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Vinod Kumar Punia
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2The numbers become $1^{200},1^{2000},1^{200000}$, which are $1,1,1$, so the limit is 1. – Empy2 Dec 03 '15 at 16:53
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You nailed it man – lcv Dec 03 '15 at 18:20
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When $x$ gets small enough (in fact, only $<1$ does the job), $[x]$ is exactly $0$ (this is very different from saying that it has a zero limit).
This allows us to say that for small enough $x$, $(1 + [x])^{2/x} = 1^{2/x} = 1$ and so has a limit $1$.
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But in $1^{\frac{2}{x}},\frac{2}{x}$ tends to infinity.Isn't it?So this is in the form of $1^{\infty}$@Ahmed Hussein – Vinod Kumar Punia Dec 03 '15 at 16:48
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1@VinodKumarPunia, the form $1^\infty$ refers to $f(x)^{g(x)}$ where $f(x)\to 1$ and $g(x)\to\infty$. When $f(x)$ is identically $1$, as in this example, it isn't an indeterminate form anymore. – vadim123 Dec 03 '15 at 17:09
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$$\lim_{x\to 0^+}\left(1+\lfloor{x}\rfloor\right)^{\frac{2}{x}}=$$ $$\lim_{x\to 0^+}\exp\left(\ln\left(\left(1+\lfloor{x}\rfloor\right)^{\frac{2}{x}}\right)\right)=$$ $$\lim_{x\to 0^+}\exp\left(\frac{2\ln\left(1+\lfloor{x}\rfloor\right)}{x}\right)=$$ $$\exp\left(\lim_{x\to 0^+}\frac{2\ln\left(1+\lfloor{x}\rfloor\right)}{x}\right)=$$ $$\exp\left(2\left(\lim_{x\to 0^+}\frac{\ln\left(1+\lfloor{x}\rfloor\right)}{x}\right)\right)=$$ $$\exp\left(2\left(0\right)\right)=\exp\left(0\right)=e^0=1$$
Jan Eerland
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