evaluate lim x^3 -8 divided by x^2 -4 as x approaches 2
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limit ((x^3-8)/(x^2-4)) as x->2
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At a 0/0 singularity such as this, you just take the ratio of derivatives:
(3x^2)/2x = (3/2)*x = 3
That is called L'Hopital's Rule.
(x-2)(x^2+2x+4)
=================
(x-2)(x+2)
(x^2+2x+4)
=================
(x+2)
when x = 2
(4+4+4)
=================
(2+2)
3
To evaluate the limit of (x^3 - 8) / (x^2 - 4) as x approaches 2, you can use direct substitution to find the value of the expression at x = 2. However, direct substitution may lead to an undefined result because it results in division by zero. Therefore, we need to factorize the expression to simplify it before evaluating the limit.
The numerator (x^3 - 8) can be written as (x - 2)(x^2 + 2x + 4), using the formula for the difference of cubes. Similarly, the denominator (x^2 - 4) can be factored as (x - 2)(x + 2).
Now, we can simplify the expression as follows:
lim as x approaches 2: [(x - 2)(x^2 + 2x + 4)] / [(x - 2)(x + 2)]
Since the (x - 2) term appears in both the numerator and denominator, we can cancel it out:
lim as x approaches 2: (x^2 + 2x + 4) / (x + 2)
Finally, we can substitute x = 2 into the simplified expression:
lim as x approaches 2: (2^2 + 2 * 2 + 4) / (2 + 2)
= (4 + 4 + 4) / 4
= 12 / 4
= 3
Therefore, the value of the expression (x^3 - 8) / (x^2 - 4) as x approaches 2 is 3.