Post

Created by @nathanedwards
 at November 3rd 2023, 9:06:37 pm.

Question:

Evaluate the following limit algebraically:

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

Answer:

To find the limit, we can't simply substitute x=2x = 2 into the expression, because it would result in an indeterminate form of 00\frac{0}{0}. Instead, we need to manipulate the expression algebraically to simplify it before taking the limit.

Let's factor the numerator and denominator of the expression:

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

Now, we can cancel out the common factor of (x−2)(x - 2) in the numerator and denominator:

lim⁡x→2(x−2)(x2+2x+4)x(x−4)=lim⁡x→2x2+2x+4x−4\lim_{x \to 2} \frac{(x - 2)(x^2 + 2x + 4)}{x(x - 4)} = \lim_{x \to 2} \frac{x^2 + 2x + 4}{x - 4}

Since the denominator still evaluates to zero at x=2x = 2, we need to simplify further. Notice that the numerator of the expression can be factored as well:

lim⁡x→2x2+2x+4x−4=lim⁡x→2(x−2)(x+2)+8x−4\lim_{x \to 2} \frac{x^2 + 2x + 4}{x - 4} = \lim_{x \to 2} \frac{(x - 2)(x + 2) + 8}{x - 4}

Now, we can cancel out the factor of (x−2)(x - 2):

lim⁡x→2(x−2)(x+2)+8x−4=lim⁡x→2x+2+8x−4\lim_{x \to 2} \frac{(x - 2)(x + 2) + 8}{x - 4} = \lim_{x \to 2} \frac{x + 2 + 8}{x - 4}

Finally, we can substitute x=2x = 2 into the simplified expression:

lim⁡x→2x+2+8x−4=2+2+82−4=12−2=−6\lim_{x \to 2} \frac{x + 2 + 8}{x - 4} = \frac{2 + 2 + 8}{2 - 4} = \frac{12}{-2} = -6

Therefore, the limit of the given expression, as xx approaches 2, is equal to −6-6.