Question

Help please, how would I find this limit?

Answer 1

\[\lim_{x \rightarrow 0+}x(\ln x)^2\]

Answer 2

separate it into product of limits, I guess.

Answer 3

looks like \(0\times -\infty\)
the gimmick here is to rewrite it as
\[\lim_{x\to 0^+}\frac{(\ln(x))^2}{\frac{1}{x}}\] the use l'hopital

Answer 4

It also says: (Hint: either substitute x = t^2 or write x = (x^1/2)^2 and combine.)

Answer 5

btw i was wrong, it is not \(0\times -\infty\) it is \(0\times \infty\) i forgot about the square

Answer 6

I graphed it on desmos, and it does seem as though it is 0
https://www.desmos.com/calculator/sowqhq9bjg

Answer 7

Do any of you guys know how to solve it using the hint, like with the substitution?

Answer 8

no

Answer 9

\[\lim_{x \rightarrow 0+}x(\ln x)^2\]
\[\lim_{t \rightarrow 0+}t^2(\ln x(t^2))^2\]

Answer 10

\[\lim_{t \rightarrow 0+}t^2(\ln (t^2))^2\]
\[\lim_{t \rightarrow 0+}t^24(\ln (t))^2\]

Answer 11

not sure what that accomplishes
why do all the limit questions come with the stupid proviso "no l'hopital"?