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

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