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

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We are given that f(x)=1+2x^5/x^2 = 1 + 2x^3.

We have to find: lim x -->1 [(f(x) - f(1))/(x-1)]

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

=> lim x -->1 [(2x^3 - 2)/(x-1)]\

=> lim x -->1 [2*(x - 1)(x^2 + x + 1)/(x-1)]

=> lim x -->1 [2*(x^2 + x + 1)]

substitute x with 1

=> 2*(1 +1 +1)

=> 6

**Therefore the required limit is 6.**

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.**