I can't find the mistake in this limit

I can't find the mistake in this limit

Postby Guest » Sun Jan 17, 2021 4:04 pm

Could you please tell me what is wrong?
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Re: I can't find the mistake in this limit

Postby nathi123 » Mon Jan 18, 2021 4:39 am

Let f(x)=[tex](\frac{1}{x})^{\frac{1}{lnx}} \Rightarrow lnf(x)=ln(\frac{1}{x})^{\frac{1}{lnx}} =\frac{1}{lnx} .ln\frac{1}{x}=\frac{ln\frac{1}{x}}{lnx}[/tex]
[tex]\lim_{x \to \infty} lnf(x)=\lim_{x \to \infty}\frac{ln\frac{1}{x}}{lnx}=\lim_{x \to \infty}\frac{\frac{1}{x}.(-\frac{1}{x^{2}})}{\frac{1}{x}}=\lim_{x \to \infty}(-\frac{1}{x^{2}})=0[/tex]. (Lopital)
[tex]\lim_{x \to \infty}f(x)=e^{0}=1[/tex].

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Re: I can't find the mistake in this limit

Postby HallsofIvy » Mon Jan 18, 2021 1:56 pm

nathi123 wrote:Let f(x)=[tex](\frac{1}{x})^{\frac{1}{lnx}} \Rightarrow lnf(x)=ln(\frac{1}{x})^{\frac{1}{lnx}} =\frac{1}{lnx} .ln\frac{1}{x}=\frac{ln\frac{1}{x}}{lnx}[/tex]
[tex]\lim_{x \to \infty} lnf(x)=\lim_{x \to \infty}\frac{ln\frac{1}{x}}{lnx}=\lim_{x \to \infty}\frac{\frac{1}{x}.(-\frac{1}{x^{2}})}{\frac{1}{x}}=\lim_{x \to \infty}(-\frac{1}{x^{2}})=0[/tex]. (Lopital)
[tex]\lim_{x \to \infty}f(x)=e^{0}=1[/tex].

[tex]\ln\left(\frac{1}{x}\right)= \ln(x^{-1})= -ln(x)[/tex]

So [tex]\frac{ln\left(\frac{1}{x}\right)}{ln(x)}= \frac{-ln(x)}{ln(x)}= -1[/tex]
for all positive x. Of course it is not defined for x non-positive.

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Re: I can't find the mistake in this limit

Postby Guest » Fri Oct 15, 2021 2:59 am

[tex]f(x)=(\frac{1}{x})^{\frac{1}{lnx}}[/tex]
[tex]ln(f(x))=ln((\frac{1}{x})^{\frac{1}{ln(x)}})=\frac{1}{ln(x)}\cdot ln(\frac{1}{x})=\frac{1}{ln(x)}\cdot ln(x^{-1})=\frac{1}{ln(x)}\cdot(-1)\cdot ln(x)=(-1)\cdot\frac{ln(x)}{ln(x)}=-1[/tex]
[tex]f(x)=e^{-1}=\frac{1}{e}[/tex]
[tex]x\in(0,1)\cup(1,+\infty)[/tex]
[tex]\lim_{x\to+\infty}f(x)=\lim_{x\to+\infty}(\frac{1}{x})^{\frac{1}{lnx}}= \lim_{x\to+\infty}\frac{1}{e}=\frac{1}{e}[/tex]
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