Evaluate the limit of the fraction (f(x)-f(1))/(x-1), if f(x)=1+2x^5/x^2? x->1

To evaluate the limit of the given fraction means
to calculate the value of the first derivative in  the given point, x =
1.

limit [f(x) - f(1)]/(x-1), when x ->
1.

First, we'll simplify f(x) =
1+2x^5/x^2

f(x) = 1 +
2x^(5-2)

f(x) = 1 + 2x^3

We'll
calculate the value of f(1):

f(1) = 1 +
2*1^3

f(1) = 1 + 2

f(1) =
3

limit [f(x) - f(1)]/(x-1) = lim (2*x^3 + 1 - 3)/(x -
1)

We'll combine like
terms:

lim (2*x^3 + 1 - 3)/(x - 1) = lim (2*x^3 - 2)/(x -
1)

We'll factorize the numerator by
2:

lim (2*x^3 - 2)/(x - 1) = lim
2(x^3-1)/(x-1)

We'll write the difference of cubes as a
product:

x^3 - 1 = (x-1)(x^2 + x +
1)

lim 2(x^3-1)/(x-1) = 2 lim (x-1)(x^2 + x +
1)/(x-1)

We'll simplify the ratio and we'll
get:

2 lim (x-1)(x^2 + x + 1)/(x-1) = 2 lim (x^2 + x +
1)

We'll substitute x by 1 and we'll
get:

2 lim (x^2 + x + 1) = 2(1^2 + 1 +
1)

2 lim (x^2 + x + 1) = 2*3

2
lim (x^2 + x + 1) = 6

But f'(x) =
f'(1)

f'(1) = limit [f(x) -
f(1)]/(x-1)

limit [f(x) - f(1)]/(x-1) = 6,
when x->1.