Question

Find the limit of t(lnt) as t approaches 0+

Answer 1

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

Answer 2

substitute..

Answer 3

seems like a l'hospital

Answer 4

\[t=\frac1n\]so as\[t\to0^+\implies n\to\infty\]then yes, l'hospital

Answer 5

though there is probably a cooler zarkon way to do it :)

Answer 6

if you sub it will be 0 x -infinity

so use l'hospital

\[-\infty \times \frac{1}{\infty} \implies \frac{-\infty}{\infty}\]

Answer 7

usual gimmick is to write as
\[\frac{\ln(x)}{\frac{1}{x}}\]

Answer 8

I would just write it as \[\lim_{x \rightarrow 0^+} \frac{\ln t}{1/t}\] and use L'hospitals rule

Answer 9

oh why do I always forget that^
:S

Answer 10

so turn it into
\[\frac{\ln t}{\frac{1}{t}}\]

Answer 11

change my x to a t

Answer 12

lol change mine as well

Answer 13

that is what i get by coping and pasting

Answer 14

or maybe use \(\xi\)

Answer 15

add \(\lim_{t \rightarrow 0^+}\) to mine

Answer 16

\[\eta\]

Answer 17

\[\large \lim_{t \rightarrow 0^+} \frac{\frac{d}{dx} \ln t}{\frac{d}{dx} \frac{1}{t}}\]

Answer 18

change that x to t

Answer 19

lol

Answer 20

\[\lim_{\diamondsuit\to 0^+}\frac{\ln(\diamondsuit)}{\frac{1}{\diamondsuit}}\]

Answer 21

nice

Answer 22

your picture here \(\diamondsuit\)

Answer 23

\[\large\lim_{\heartsuit \rightarrow 0^+} \frac{\ln (\heartsuit)}{\frac{1}{\heartsuit}}\]

Answer 24

@davidesm you there? lol

Answer 25

I apologize, but how did you get from the original question to lnt/1/t?

Answer 26

look at the 6th post,,maybe that'll help

Answer 27

multiplying by \(t\) is the same as dividing by the reciprocal of \(t\)
just like multiplying by 2 is the same as dividing my \(\frac{1}{2}\)
it is just an algebra trick to turn it in to \(\frac{-\infty}{\infty}\) instead of \(0\times (-\infty)\)