Question

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

Answer 1

`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`