lim (x->3) square root of ((8x^3-9)/(4x^2-9))Limit as x approaches 3 / 2 of the square root of 8 x to the 3 minus 27,divided by 4 x squared minus 9

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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lim (8x^3 - 27)/(4x^2 - 9)    x --> 3/2

First let us substitute:

lim =  (8*27/8 - 27)/(4*9/4 - 9) = 0/0

The method failed because 3/2 is a root for both numerator and denominator.

Let us facto:

= lim (2x-3)(4x^2 + 6x + 9)/(2x-3)(2x+3)

Reduce similars:

= lim (4x^2 + 6x + 9)/(2x+3)   x--> 3/2

==> lim as x--> 3/2 = (4*9/4 + 6*3/2 + 9)/(2*3/2 +3)

                                  = 27/6 = 9/2

Then the limit is 9/2

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

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To determine lt sqrt((8x^3-9)/(4x^2-9)) as x approaches 3,

we first  substitute x = 3 in  sqrt ((x^3-8)/(4x^2-9))  and we find :

sqrt (( 8*3^3-9)/(4*3^2-9)) = sqrt(207/27) = sqrt(23/3) =2.7689 nearly.

(ii)

Lt as x-->3/2  of  sqrt((8x^3-27)/(4x^2-9)) = Lt{[8*(3/2)^3-27]/{[(3/2)^2-9]} =  sqrt(27-27)/(9-9) indeterminate form.

As x= 3/2  makes both numerator and denominators zero, (x-3/2) or 2x-3  is factor of both numerator denominator. So we divide both numerator and denominator by 2x-3 inside  the square root.and then take the limit .

(8x^3-27)/(2x-3) = 4x^2 +6x+9,

(4x^2-9)/(2x-9) = 2x^2+3.

Therefore  Lt sqrt((8x^3-27)/(4x^2-9)) = lt sqrt((4x^2+66x+9)/(2x+3)) =sqrt ((4(3/2)^2+6(3/2)+9)/(2(3/2)+3) = sqrt((9+9+9)/(3+3)) = sqrt 4.5 = sqrt(9/2) = (3sqrt2)/2

 

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