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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justaguide's profile pic

Posted on

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.

giorgiana1976's profile pic

Posted on

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.

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