Question

Evaluate the limit as x goes to 0+ (x^2*lnx) using l'hopitals rule

Answer 1

\[\lim_{x\to0^+}x^2\ln x\]?

Answer 2

yes

Answer 3

I took the derivative using the product rule which was (x^2*1/x)+(2x*lnx)

Answer 4

to use l'Hospital, fist we need to convert it into the form \(\frac00\)

I suppose we can do the transformation \(x=\frac1t\) which implies that \(t\to\infty\) as \(x\to0^+\)

Answer 5

l'hospital requires us to take the derivative of the numerator and denominator separately, not just take the derivative of the whole function like you did

Answer 6

but does that mean i would need to change the form of the equation to have a denominator?

Answer 7

yes, try the substitution \(x=\frac1t\)

note that will change your limit though, because since we want x to approach 0, than implies that t must go to infinity

Answer 8

ok

Answer 9

so if t goes to infinity using that substitution it becomes 1 over infinity which is 0

Answer 10

x goes to zero, yes
and what does the whole expression become?

Answer 11

0/0?

Answer 12

just sub x=1/t for now
no limit taking yet

Answer 13

well x^2lnx=lnx/1/x^2
so using this we'll have:
lim as x ->0+ lnx/1/x^2=lim as x ->0+ 1/x/(-2/x^3)=lim as x ->0+ -x^2/2=0

Answer 14

\[\lim_{x \rightarrow 0^+}x^2lnx=\lim_{x \rightarrow 0^+} lnx/(1/x^2)=\lim_{x \rightarrow 0^+} 1/x/(-2/x^3)\]
\[=\lim_{x \rightarrow 0^+} -x^2/2=0\]

Answer 15

ok, I'll give it to @anonymoustwo44
his/her way is better than mine. they both give the right answer though :)

Answer 16

yep :D

Answer 17

thank you both of you i just seemed to be confused with the natural log :/

Answer 18

yours is unique @TuringTest

Answer 19

naw, I use the \(x=\frac1t\) trick all the time
so does wolfram actually :P

Answer 20

does this mean that every time it is an indeterminate form the limit will be 0?

Answer 21

no

Answer 22

indeterminate means what it sounds like:
the limit is yet to be determined

Answer 23

mmmm okay

Answer 24

can you guys help me with one more problem? please

Answer 25

sure, could you please post separately?

Answer 26

ok

Answer 27

it is the same using l'hopitals rule
limit -->1+
(x^2-1)^(x-1)

Answer 28

\[\lim_{x \rightarrow 1} \ln(x^2)/(x^2-1)=1?\]

Answer 29

is this correct?

Answer 30

Hi, Jenjen. Welcome to Open Study =D
It's considered more polite to close this question and then post a new question rather than asking in a comment.

Answer 31

thanks for the welcome lol ill do that ><