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I have the following limit:

$$\lim_{x\rightarrow0^+}{-x\log{x}}$$

Since this leads to an indeterminate form $[0\cdot\infty]$, I change the variables:

$$t=\frac 1 x$$

$$\lim_{t\rightarrow+\infty}-{\frac{\log(\frac{1}{t})}{t}}$$

The denominator grows faster than the nominator so the fraction tends to $0^+$. But since there is a minus there, I flip the value which becomes $0^-$. According to my textbook this is wrong. The limit is supposed to be $0^+$. Any hints on why my approach to get the $\pm$ is wrong?

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You are wrong since $$-\log{1/x}=\log x$$therefore $$\lim_{t\to\infty}\dfrac{\log t}{t}=0^+$$

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  • $\begingroup$ It doesn't matter but since the question has asked for whether the limit is $0^+$ or $0^-$ we mention it here $\endgroup$ – Mostafa Ayaz Aug 21 '18 at 8:30
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Hint:

$$\log\left(\frac1t\right)=-\log t$$

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  • $\begingroup$ That's right, thanks $\endgroup$ – Cesare Aug 21 '18 at 8:25
  • $\begingroup$ You are welcome off course. $\endgroup$ – drhab Aug 21 '18 at 8:25

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