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.