What is $lim_(x->0) xlnx - x^2$ ?

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Answer 1 · 2015-04-10T02:29:26

$y = xlnx - x^2$

If you merely plug in zero, what you get is undefined.

$lim_(x->0) f(x) =$ undefined by simply using $x = 0$ .

You can try this:

$y = lnx/(1/x) - x^2$
$= lnx/(1/x) - x/(1/x)$
$= (lnx - x)/(1/x)$

Use L'Hopital's Rule to differentiate this (separately for numerator and denominator), because now it has a nice form for it.

$y = (1/x - 1)/(-1/x^2)$
$= (-x^2)(1/x - 1)$
$= -x + x^2$

Now, set them equal to $0$ .

$x^2 - x = 0$

So, this limit is equal to zero.

graph{xlnx - x^2 [-1.552, 1.866, -1.019, 0.69]}

What is lim_(x->0) xlnx - x^2lim_(x->0) xlnx - x^2?