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The first choice is correct: the limit is 0.
The multiplication rule for the limits cannot be applied here because, while the limit of `x^2` as x approaches `0^+` exists and equal to 0, the limit of ln(x) does not exist as x approaches `0^+` (0 from the right). As can be seen on the graph below, the function ln(x) is unbounded and approaches negative infinity when x approaches `0^+` :
Then, the behavior of the function `f(x) = x^2ln(x)` close to 0 has to be analyzed in order to determine the limit. It can be seen from the graph of f(x) that the function will approach zero:
One way to prove this is by using L'Hospital rule. Rewrite f(x) as
`f(x) = lnx/(1/x^2)`
As x approaches 0, both numerator and denominator approach infinity. Take the derivative of the numerator and denominator:
`(ln(x))' = 1/x`
`(1/x^2)' = -2/x^3`
According to the L'Hospital rule, the limit of the fraction will equal the limit of the ratio of the derivatives, which is
`(1/x)/((-2)/x^3) = x^3/(-2x) = x^2`
The limit of x^2 is 0 as x approaches 0, so the limit of f(x) is also 0.
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