Question

Evaluate the limit below... (i will type it)

Answer 1

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

Answer 2

\[xln(x)\]

Answer 3

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

Answer 4

direct substitution limit law, just put x=0. even though you are only looking at the right side limit, if you find the limit, then the left and right side limits are equal to that. im pretty sure this is correct

Answer 5

that just gives undefined, I think you may need the delta-epsilon thing

Answer 6

@colton: makes no sense here whatsoever because ln x doesn't exist when x = 0.

Answer 7

lHospitals rule

Answer 8

very easy

Answer 9

i know its a inderterminate form of the type \[(0 * -\infty)\] And need to use L'Hospital's rule i believe??

Answer 10

you can use L'hospital on this? I thought it had to evaluate to 0/0 to use L'hospital's rule

Answer 11

oh yeah, sorry, my bad

Answer 12

yes, following Zarkon's lead, write

x ln x = ln x / (1/x)

Answer 13

ahhh!

Answer 14

\[\frac{\ln(x)}{1/x}\]

Answer 15

My professor, in some of his examples, manipulated the equation in such a way that it ended up to be 0/0 or infinity/infinity but im not sure how to do that?

Answer 16

Zarkon and James just did it, now use the rule on what Zarkon typed

Answer 17

and then you would take the derivative of the top and bottom \[\frac{\frac{1}{x}}{-\frac{1}{x^2}}\]

Answer 18

right which gives -x

Answer 19

ahh i see. Which would be 0/-infinity.. which is 0.. Now why the hell couldnt i figure that out on the test >.< lol.. Ty very much