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...
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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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