What is the limit of the function [f(x)-f(1)]/(x-1), x-->1, if f(x)=square root[(2x+3)^3]/3?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The function f(x) = sqrt[(2x+3)^3]/3

We have to find the value of lim x-->1 [(f(x) - f(1))/(x - 1)]

lim x-->1 [(f(x) - f(1))/(x - 1)]

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

=> lim x-->1 [(sqrt[(2x+3)^3]/3 - sqrt[5^3]/3)/(x - 1)]

If we substitute x = 1, we get the indeterminate form 0/0. Therefore we can use l'Hopital's rule and replace the denominator and numerator with their derivatives

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

substitute x = 1

=> (1/2)(1/3)*3*(2 + 3)^2*2/(sqrt[(2+3)^3])

=> (1/2)(1/3)*3*5^2*2/sqrt[5^3]

=> 25/sqrt 125

=> 25/5*sqrt 5

=> sqrt 5

The required limit is sqrt 5

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We notice that the limit of the fraction is the derivative of f(x) function at the point x = 1.

Therefore,

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

We'll have to determine the 1st derivative of the given function f(x):

f'(x) = [(2x+3)^(3/2)]/3

We'll apply chain rule to determine f'(x):

f'(x) = 2*(3/2)[(2x+3)^(1/2)]/3

We'll simplify and we'll get:

f'(x) = sqrt(2x+3)

Now, we'll determine f'(1):

f'(1) = sqrt(2+3)

f'(1) = sqrt5

The requested limit of the fraction [f(x)-f(1)]/(x-1), if x approaches to 1, is lim [f(x)-f(1)]/(x-1) = sqrt 5.

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