Please explain and evaluate each of the following limits and show all the work

Calculus
Tutor: None Selected Time limit: 1 Day

Jun 14th, 2015

Thank you for the opportunity to help you with your question!

so given f(x) = x - 4x^2

now f(x+h) = (x+h)  - 4(x+h)^2

                 =  (x+h) - 4(x^2 + h^2 + 2xh)

                = (x+h) -4x^2 - 4h^2 -8xh

so now limit h -->0  f(x+h) - f(x) / h   =  [  (x+h) - 4x^2 -4h^2 -8xh  - x +4x^2 ] / h

                                                           = [ h -4h^2 -8xh ] / h

                                                           = h [ 1 -4h -8x ] /h

                                                          = [ 1 -  4h -8x ]

so but h ->0 so plug h value

now we get =  1 - 4*0 -8x

                   = 1 - 8x


for the second one post it in another question I will answer thank you. comment for any doubts

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 14th, 2015

Ok I will answer here it self , But pay me  $1 later. thanks

Jun 14th, 2015

okay i will pay you a dollar, are you answering the second one here too?

Jun 14th, 2015

okay i will pay you a dollar, are you answering the second one here too?

Jun 14th, 2015

how does at the end 1-4*0-8x ends up being 1-8x

Jun 14th, 2015

how does at the end 1-4*0-8x ends up being 1-8x

Jun 14th, 2015

 it is  4 * zero = zero

something into zero becomes zero

Jun 14th, 2015

given x ->0  [ 1/r + 1/x-r ] / x

so for your understanding  first take  1/r  + 1/x-r  and simplify

1/r  + 1/x-r  =  [ x-r +r ] / (r) * (x-r)

  =  x / (r) * (x-r)

now plug it in the actual problem  x -->0  x / (r) * (x-r) *x

in the numerator and denominator we have x , so both will get cancel

 x ->0  [ 1/r + 1/x-r ] / x  =  1 / (r) * (x-r)

so now plug x =0

we will get  =  1 / (r) * (0 -r)

  =  1 / ( r) * (-r)

  = - 1 /  (r^2)


Jun 14th, 2015

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