So, we'll have to calculate 2
limits.
1) First, we'll calculate the limit of the
expression:
sqrt [(8x^3-9)/(4x^2-9)], when x->
3
To calculate the limit, we'll have to substitute x by the
indicated value, namely 3. We'll check if we'll get an indeterminacy
case.
lim sqrt [(8x^3-9)/(4x^2-9)] =
sqrt[(8*3^3-9)/(4*3^2-9)]
lim sqrt [(8x^3-9)/(4x^2-9)] =
sqrt 207/27
lim sqrt [(8x^3-9)/(4x^2-9)] = sqrt
7.66
lim sqrt [(8x^3-9)/(4x^2-9)] = 2.76
approx.
2) lim sqrt [(8x^3-27)/(4x^2-9)], if
x - > 3/2
To calculate the limit, we'll have to
substitute x by the indicated value, namely 3/2. We'll check if we'll get an
indeterminacy case.
lim sqrt [(8x^3-27)/(4x^2-9)] = sqrt
(8*27/8 - 27)/(4*9/4- 9)
lim sqrt [(8x^3-27)/(4x^2-9)] =
sqrt (27-27)/(9-9)
lim
sqrt [(8x^3-27)/(4x^2-9)] = 0/0,
indetermination
To calculate the limit
we'll use factorization. We notice that the numerator is a difference of
cubes:
8x^3-27 = (2x)^3 -
(3)^3
We'll apply the
formula:
a^3 - b^3 = (a-b)(a^2 + ab +
b^2)
a = 2x and b = 3
(2x)^3 -
(3)^3 = (2x-3)(4x^2 + 6x + 9)
We also notice that the
denominator is a difference of squares:
4x^2-9 = (2x)^2 -
3^2
We'll apply the
formula:
a^2 - b^2 =
(a-b)(a+b)
(2x)^2 - 3^2 =
(2x-3)(2x+3)
We'll substitute the differences by their
products:
lim sqrt [(8x^3-27)/(4x^2-9)] = lim
sqrt (2x-3)(4x^2 + 6x + 9)/(2x-3)(2x+3)]
We'll
simplify:
lim sqrt [(8x^3-27)/(4x^2-9)] = lim sqrt [(4x^2 +
6x + 9)/(2x+3)]
Now, we'll substitute x by
3/2:
lim sqrt [(4x^2 + 6x + 9)/(2x+3)] = sqrt(4*9/4 + 6*3/2
+ 9)/(2*3/2 + 3)
lim sqrt [(4x^2 + 6x + 9)/(2x+3)] = sqrt
(9+9+9)/(6)
lim sqrt [(4x^2 + 6x + 9)/(2x+3)] = sqrt
27/6
lim sqrt [(4x^2 + 6x + 9)/(2x+3)] =
3sqrt18/6
lim sqrt [(4x^2 + 6x + 9)/(2x+3)] =
3sqrt2/2
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