What is #\lim _ { x \rightarrow 0^ { + } } \sqrt { x} \ln ( x ) #?

1 Answer
256
Jan 16, 2017

#lim_(xto0^+)sqrt(x)ln(x)=0#

Explanation:

Because #ln(x)to-infty# as #xto0^+#, x goes to zero from the right, we end up with an indeterminate form.

So, we have to manipulate it algebraically

Since #x=1/(1/x)#, we can say

#sqrt(x)ln(x)=ln(x)/(1/sqrt(x))=ln(x)/x^(-1/2)#

Then we can use L'Hopital's Rule

#lim_(xto0^+)sqrt(x)ln(x)=lim_(xto0^+)ln(x)/(x^(-1/2))=^(H)lim_(xto0^+)(1/x)/(-1/2x^(-3/2))#
#=lim_(xto0^+)(-2x^(3/2))/(x)=lim_(xto0^+)-2x^(1/2)=-2(0)=0#

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