Post

Limit of Rational Function at x=2.

Created by @nathanedwards
 at November 1st 2023, 6:44:11 am.
AP_Calculus_AB-Questions
#factoring
#limits
#algebraic-simplification

Question:

Find the limit algebraically:

lim⁡x→23x2−4x−8x3−2x2−3x+2\lim_{{x \to 2}} \frac{{3x^2 - 4x - 8}}{{x^3 - 2x^2 - 3x + 2}}

Provide a step-by-step detailed explanation for your answer.

Answer:

To find the limit algebraically, we can try to simplify the expression by factoring both the numerator and denominator.

We start by factoring the numerator:

3x2−4x−8=(x+2)(3x−8)3x^2 - 4x - 8 = (x + 2)(3x - 8)

Next, we factor the denominator:

x3−2x2−3x+2=(x−1)(x−1)(x+2)x^3 - 2x^2 - 3x + 2 = (x - 1)(x - 1)(x + 2)

Now, we can rewrite the expression in terms of the factored form:

lim⁡x→2(x+2)(3x−8)(x−1)(x−1)(x+2)\lim_{{x \to 2}} \frac{{(x + 2)(3x - 8)}}{{(x - 1)(x - 1)(x + 2)}}

Notice that the factor (x+2)(x + 2) appears in both the numerator and denominator. We can cancel the common factor:

lim⁡x→23x−8(x−1)(x−1)\lim_{{x \to 2}} \frac{{3x - 8}}{{(x - 1)(x - 1)}}

At this point, we have canceled all common factors, and we can proceed to evaluate the limit by direct substitution:

lim⁡x→23x−8(x−1)(x−1)=3(2)−8(2−1)(2−1)=6−81⋅1=−21=−2\lim_{{x \to 2}} \frac{{3x - 8}}{{(x - 1)(x - 1)}} = \frac{{3(2) - 8}}{{(2 - 1)(2 - 1)}} = \frac{{6 - 8}}{{1 \cdot 1}} = \frac{{-2}}{{1}} = -2

Thus, the limit of the given expression as xx approaches 2 is -2.

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