Question

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

Accepted Answer

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.

Answer

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.

Calculus

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

Expert Answers

Unlock This Answer Now

Related Questions

Verify if limit of ln(1+x)/x is 1, x-->0

Evaluate the limit of the function (2x-sin2x)/x^3; x-->0.

How to evaluate the limit of (cos x - cos 3x) / x*sin x if x-->0 ?

Evaluate the limit of y=(2x^2-x-3)/(x+1), x-->-1 using L'Hopital theorem.

Calculate the limit of the fraction (f(x)-f(1))/(x-1) if f(x)=x^300+x+1, x-->1