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Indeterminate form problems

  • Thread starter blu3jam
  • Start date
7
0
Hi I'm new to this forum, i need a help i don't know how can i use l'hospital in these problems;


lim(x->1, ((arccos(x))^lnx))

lim(x->[tex]\pi[/tex]/2+ (cosx ln(x - [tex]\pi[/tex]/2 ) )


Can anyone help?
 
Last edited:

Answers and Replies

nicksauce
Science Advisor
Homework Helper
1,272
5
The second one is greatly simplified if you make a change of variables u = x - pi/2
 
7
0
The second one is greatly simplified if you make a change of variables u = x - pi/2

Thanks, but still i don't know how can i proceed ? =(

lim(u->0+ (cos(u+[tex]\pi[/tex]/2) ln(u ) )
 
nicksauce
Science Advisor
Homework Helper
1,272
5
Well cos(u+pi/2) = -sin(u)

Then you can write your function as
-ln(u) / csc(u), and then apply l'hopital's rule. See where that gets you.
 
7
0
Well cos(u+pi/2) = -sin(u)

Then you can write your function as
-ln(u) / csc(u), and then apply l'hopital's rule. See where that gets you.
Now ,i figured it out thank you :wink:
 

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