limit 1/(x^(1/4)ln(x)) as x approaches 0 from the right using L'Hopital's Rule

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OpenStudy (john_es):

Do you mean,
\[\frac{1}{x^{1/4}\ln x}\]

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

\[\lim_{x \rightarrow 0+} \frac{ 1 }{ x ^{1/4} \ln (x)}\]

OpenStudy (john_es):

Then transform it in infinity/infinity
\[\lim_{x\rightarrow 0^+}\frac{x^{-1/4}}{\ln x}\]Then apply L'Hopital rule,
\[\lim_{x\rightarrow 0^+}\frac{(-1/4)x^{-5/4}}{1/x}=\lim_{x\rightarrow 0^+}-x^{-1/4}/4=-\infty\]

OpenStudy (anonymous):

how did you get -x^(-1/4)/4?

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OpenStudy (anonymous):

@John_ES

zepdrix (zepdrix):

this one? :o

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

I dont know how John got the answer

OpenStudy (john_es):

Use power properties,
\[1/x=x^{-1}\]
\[\frac{x^{-5/4}}{x^{-1}}=x^{-5/4-(-1)}=x^{-1/4}\]

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OpenStudy (anonymous):

okay but if you sub in something greater than 0 you get a negative number not negative infinity.

OpenStudy (john_es):

You obtain negative infinite because the formula has a minus sign.