Post

Created by @nathanedwards
 at October 31st 2023, 5:09:18 am.

Question:

Find the limit algebraically:

lim⁡x→3x2−5x+6x−3\lim_{x \to 3}\frac{x^2 - 5x + 6}{x - 3}

Answer:

To find the limit algebraically, we can use direct substitution and simplify the expression. However, direct substitution gives us an indeterminate form (00\frac{0}{0}) since both the numerator and denominator become zero when x=3x = 3.

To resolve this indeterminate form, we can factorize the numerator and then cancel out the common factor with the denominator. Let's factorize the numerator:

x2−5x+6=(x−2)(x−3)x^2 - 5x + 6 = (x - 2)(x - 3)

Now, we can rewrite the expression:

lim⁡x→3x2−5x+6x−3=lim⁡x→3(x−2)(x−3)x−3\lim_{x \to 3}\frac{x^2 - 5x + 6}{x - 3} = \lim_{x \to 3}\frac{(x - 2)(x - 3)}{x - 3}

Notice that the factor (x−3)(x - 3) appears in both the numerator and the denominator. We can cancel out this common factor:

lim⁡x→3(x−2)\lim_{x \to 3}(x - 2)

Now we can substitute x=3x = 3 into the expression:

lim⁡x→3(x−2)=3−2\lim_{x \to 3}(x - 2) = 3 - 2

Therefore, the limit algebraically is:

lim⁡x→3x2−5x+6x−3=1\lim_{x \to 3}\frac{x^2 - 5x + 6}{x - 3} = 1