lim sqrt [( 27x^3 - 1 )/( 9x^2 - 1 )], if x -
> 1/3
To calculate the limit, we'll have to
substitute x by the indicated value, namely 1/3. We'll check if we'll get an
indeterminacy case.
lim sqrt [(27x^3 - 1)/(9x^2 - 1)] =
sqrt (27*1/27 - 1)/(9*1/9- 1)
lim sqrt [(27x^3 - 1)/(9x^2 -
1)] = sqrt (1-1)/(1-1)
lim sqrt [(27x^3 -
1)/(9x^2 - 1)] = 0/0, indetermination
To
calculate the limit we'll use factorization. We notice that the numerator is a
difference of cubes:
27x^3 - 1 = (3x)^3 -
(1)^3
We'll apply the
formula:
a^3 - b^3 = (a-b)(a^2 + ab +
b^2)
a = 3x and b = 1
(3x)^3 -
(1)^3 = (3x-1)(9x^2 + 3x + 1)
We also notice that the
denominator is a difference of squares:
9x^2-1 = (3x)^2 -
1^2
We'll apply the
formula:
a^2 - b^2 =
(a-b)(a+b)
(3x)^2 - 1^2 =
(3x-1)(3x+1)
We'll substitute the differences by their
products:
lim sqrt [(27x^3 - 1)/(9x^2 - 1)] = lim
sqrt [(3x-1)(9x^2 + 3x + 1)/(3x-1)(3x+1)]
We'll
simplify:
lim sqrt [(27x^3 - 1)/(9x^2 - 1)] = lim sqrt
[(9x^2 + 3x + 1)/(3x+1)]
Now, we'll substitute x by
1/3:
lim sqrt [(9x^2 + 3x + 1)/(3x+1)] = sqrt(9*1/9 + 3*1/3
+ 1)/(3*1/3 + 1)
lim sqrt [(9x^2 + 3x + 1)/(3x+1)] = sqrt
(1+1+1)/(1+1)
lim sqrt [(9x^2 + 3x + 1)/(3x+1)] = sqrt
(3/2)
lim sqrt [(9x^2 + 3x + 1)/(3x+1)] =
sqrt6/2
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