Can someone give me some hints on the solution of the following limit problems.
lim x to 0{cub rt(1+sinx)- cub rt(1-sinx)}/x
lim x to 0 {sin-1 x-tan-1 x}/x3
lim x to 0 { cos-1(1-x)}/x
can you please give the answers for the problems
Answers are as follows:
1) 2/3 2) Sec 2 x 3) 1/2
correct answers are 1. 2/3 2. 1/2 3. 1/2
can you elaborate?
ans for
lim x to 0{cub rt(1+sinx)- cub rt(1-sinx)}/x
it is in 0/0 form , so use l'hospital rule . u will get the ans.
lim x to 0 { cos-1(1-x)}/x
it is also in 0/0 form use l'hosp rule.
what is i'hospltal rule?
You can do these problem without L' hospital rule (which states that whenever there is 0/0 or infinity/infinity form differentiate numerator and denominator separately until you get rid of indeterminate form)
1) Muliply by (a2+b2+ab) where a= (1+sinx)1/3 and a= (1-sinx)1/3 . In numerator then you will get 2sinx then you can proceed.
2) Convert sin-1x to tan-1 form.
2) Convert cos-1 to sin-1 form.