Question

Why is the
lim x->0+ of (1/x)/(-1/x^2)
= lim x->0+ (-x)
= 0

Makes no sense to me! Why is the
lim x->0+ of (1/x)/(-1/x^2)
= lim x->0+ (-x)
= 0

Makes no sense to me! @Mathematics

Answer 1

limx->0+ of (1/x) is infinity
limx->0+ of (-1/x^2) is -infinity

So WHY does infinity/-infinity = (-x) or 0 now....

Answer 2

Makes no goddam sense...

Answer 3

If:\[x\neq 0\]then:
\[\frac{\frac{1}{x}}{-\frac{1}{x^2}}\cdot \frac{x^2}{x^2}=\frac{x}{-1}=-x\]

Answer 4

what... how am i supposed to know to do that!!!

Answer 5

you have this big fraction with small fractions in the numerator and denominator. Any time you have that, you almost always want to get rid of the internal fractions.

Answer 6

Okay, will keep that in mind! Thanks for the help!

Answer 7

Also note, its only correct to think about dividing limits if the limits actually exist. So because both:\[\lim_{x\rightarrow0}\frac{1}{x}\]and \[\lim_{x\rightarrow 0}\frac{1}{x^2}\]dont exist, try not to think about division.

Answer 8

Well i'm doing l'hopitals, so I have to think about division, right?. That was just one step from original function xlnx in my book that I did not understand. They didn't mention at all that they multiplied by x^2 so I was very confused.

Answer 9

ah, i see. yeah, youre right.

Answer 10

Joe, I have one more quick question. I feel they are going to trick me tomorrow on the quiz giving me a question that looks like it needs L'Hoptals, but actually doesn't.

The first conditions for lhopitals is that f and g be differentiable. Which implies continuity. Which I understand. But then my book says that \[g'(x)\neq0\] near a. Shouldnt it be \[f'(x)\neq0 AND g'(x)\neq0\] Why did they only mention g'(x)?

Answer 11

I know all the intermediate form types, so don't worry about that. It's just that one condition I'm a bit unclear on.