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First of all I tried to solve this limit problem which says:
limit x->0 (x^2)*ln(1/x)
What I've tried is limit x->0 -x^2*lnx, but I couldn't complete since plugging x is 0*0 = 0 so the limit is 0 or what?

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    $\begingroup$ Hint: Let $x = e^u$. $\endgroup$ Dec 26, 2017 at 16:55
  • $\begingroup$ If $x$ tends to $0$ (from the right), $\ln(1/x)$ tends to $\infty$, so you cannot just plug in $x=0$. And $\ln(x)$ tends to $-\infty$; the same problem. $\endgroup$
    – Peter
    Dec 26, 2017 at 16:56

3 Answers 3

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Let $x_n=e^{-n}$, which tends to $0$. Then

$$t_n=ne^{-2n}$$ and $$\frac{t_{n+1}}{t_n}=\frac{n+1}{ne^2}<\frac2{e^2}$$

so that the sequence converges to $0$.

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$$\lim _{ x\rightarrow 0 }{ \left[ { x }^{ 2 }\ln { \left( \frac { 1 }{ x } \right) } \right] } =-\lim _{ x\rightarrow 0 }{ \frac { \ln { x } }{ \frac { 1 }{ { x }^{ 2 } } } } \overset { L'Hospital's }{ = } -\lim _{ x\rightarrow 0 }{ \frac { \frac { 1 }{ x } }{ \frac { -2 }{ { x }^{ 3 } } } } =\lim _{ x\rightarrow 0 }{ \frac { { x }^{ 2 } }{ 2 } =0 } $$

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The limit is 0. Use L'hôpital rule to find limits of such forms as this. Convert this into a 0/0 form first by taking the x^2 to the denominator.

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