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

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

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

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

it is 4 * zero = zero

something into zero becomes zero

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)

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