Limit xam tends to 2 [x-2/x^2-x-1/x^3-3x^2+2x]
Answers (2)
6th Apr, 2019
Hellooooo.......
Here below i am providing solution to your problem by expecting your question be like ....
Lim x-->2 (x-2)(x^2-x-1)/(x^3-3x^2+2x)
By applying L hospital rule differentiating on numerator and denominator we get
Limx-->2 (x-2)(2x-1)+(x^2-x-1)(1)/(3x^2-6x+2)
By substituting we get
1/2
As solution...
If your question was something different mention below in comment box....
Thank you..
Here below i am providing solution to your problem by expecting your question be like ....
Lim x-->2 (x-2)(x^2-x-1)/(x^3-3x^2+2x)
By applying L hospital rule differentiating on numerator and denominator we get
Limx-->2 (x-2)(2x-1)+(x^2-x-1)(1)/(3x^2-6x+2)
By substituting we get
1/2
As solution...
If your question was something different mention below in comment box....
Thank you..
1 like
Comments (0)
6th Apr, 2019
Hello
please write the question properly using brackets.
And divisions bis thrice. Give it in a p/q format.
Thank you
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