Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
frx
Group Title
Show, using L'Hopitals rule, that
\[\lim_{x \rightarrow 0} x^{a}lnx = 0\]
for a>0
I started by doing the first derivitive but it didn't help much and doing more derivitives just made it worse, how should i continue?
\[\lim_{x \rightarrow 0} \frac{ d }{dx} x ^{a}lnx = \lim_{x \rightarrow 0} x^{a1}(alnx+1)\]
 one year ago
 one year ago
frx Group Title
Show, using L'Hopitals rule, that \[\lim_{x \rightarrow 0} x^{a}lnx = 0\] for a>0 I started by doing the first derivitive but it didn't help much and doing more derivitives just made it worse, how should i continue? \[\lim_{x \rightarrow 0} \frac{ d }{dx} x ^{a}lnx = \lim_{x \rightarrow 0} x^{a1}(alnx+1)\]
 one year ago
 one year ago

This Question is Closed

alexandercpark Group TitleBest ResponseYou've already chosen the best response.0
remember that you can only apply l'hospital rule on specific forms (the only one i remember i 0/0 at the moment.) remember that \[x^a = \ln (e ^{x^a})\]
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
I forgot that, so now I should just derivate? So: \[\lim_{x \rightarrow 0} \frac{ d }{ dx} \ln(e ^{x ^{a}})lnx=\lim_{x \rightarrow 0}\frac{ ax ^{a} lnx \ln(e ^{x ^{a}}) }{ x } \] And that is 0/0, and to get the answer should I just continue to derivate?
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
It does get really messy to derivate more.. or maybe I shouldn't derivate further?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
dw:1351678852308:dw
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
Ohh i see, thanks :)
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.