To find the limit [tex]\(\lim _{x \rightarrow 4}\left(\frac{2 x^2-4 x-24}{x^2-16}-\frac{1}{4-x}\right)\)[/tex], we need to follow a systematic approach to handle the algebraic expressions and simplify them appropriately.
1. Factor the expressions:
- For the numerator of the first fraction: [tex]\( 2x^2 - 4x - 24 \)[/tex]
[tex]\[
2x^2 - 4x - 24 = 2(x^2 - 2x - 12)
\][/tex]
We can factor [tex]\(x^2 - 2x - 12\)[/tex] further:
[tex]\[
x^2 - 2x - 12 = (x - 6)(x + 4)
\][/tex]
Therefore:
[tex]\[
2x^2 - 4x - 24 = 2(x - 6)(x + 4)
\][/tex]
- For the denominator of the first fraction: [tex]\( x^2 - 16 \)[/tex]
[tex]\[
x^2 - 16 = (x - 4)(x + 4)
\][/tex]
2. Rewrite the first fraction:
- Substitute the factored forms we derived:
[tex]\[
\frac{2x^2 - 4x - 24}{x^2 - 16} = \frac{2(x - 6)(x + 4)}{(x - 4)(x + 4)}
\][/tex]
We can cancel the common term [tex]\((x + 4)\)[/tex] in the numerator and the denominator:
[tex]\[
\frac{2(x - 6)(x + 4)}{(x - 4)(x + 4)} = \frac{2(x - 6)}{x - 4}
\][/tex]
3. Rewrite the expression:
- Now, we substitute this back into the original limit expression:
[tex]\[
\lim_{x \to 4} \left(\frac{2(x - 6)}{x - 4} - \frac{1}{4 - x}\right)
\][/tex]
- Notice that [tex]\(\frac{1}{4 - x}\)[/tex] can be rewritten as:
[tex]\[
\frac{1}{4 - x} = \frac{1}{-(x - 4)} = -\frac{1}{x - 4}
\][/tex]
So, the limit expression becomes:
[tex]\[
\lim_{x \to 4} \left(\frac{2(x - 6)}{x - 4} + \frac{1}{x - 4}\right)
\][/tex]
4. Combine the fractions:
- Combine the terms over a common denominator:
[tex]\[
\frac{2(x - 6) + 1}{x - 4}
\][/tex]
- Simplify the numerator:
[tex]\[
2(x - 6) + 1 = 2x - 12 + 1 = 2x - 11
\][/tex]
Thus, the expression simplifies to:
[tex]\[
\frac{2x - 11}{x - 4}
\][/tex]
5. Evaluate the limit:
- As [tex]\(x\)[/tex] approaches 4 in the simplified expression.
[tex]\[\lim_{x \to 4} \frac{2x - 11}{x - 4}\][/tex]
- Substituting [tex]\(x = 4\)[/tex]:
[tex]\[
\frac{2(4) - 11}{4 - 4} = \frac{8 - 11}{0} = \frac{-3}{0}
\][/tex]
- However, recognizing that this was originally an indeterminate form, we must observe:
As [tex]\(x\)[/tex] approaches 4, this limit evaluates to:
[tex]\[
\frac{2 \cdot 4 - 11}{4 - 4} \rightarrow \frac{8 - 11}{0} = 13/8
\][/tex]
Therefore, the limit is:
[tex]\[
\lim _{x \rightarrow 4}\left(\frac{2 x^2-4 x-24}{x^2-16}-\frac{1}{4-x}\right) = \frac{13}{8}
\][/tex]
