Let's solve the following two problems step-by-step.
### Problem 1:
Simplify the expression [tex]\(\frac{x^2 - 36}{x - 6}\)[/tex]
Solution:
First, notice that the numerator [tex]\(x^2 - 36\)[/tex] can be factored using the difference of squares formula: [tex]\(a^2 - b^2 = (a - b)(a + b)\)[/tex].
In this case, we have [tex]\( x^2 - 36 = (x - 6)(x + 6) \)[/tex].
So we can rewrite the expression as:
[tex]\[
\frac{x^2 - 36}{x - 6} = \frac{(x - 6)(x + 6)}{x - 6}
\][/tex]
If [tex]\(x \neq 6\)[/tex], the [tex]\( x - 6 \)[/tex] terms cancel each other out, leaving:
[tex]\[
\frac{(x - 6)(x + 6)}{x - 6} = x + 6
\][/tex]
Therefore,
[tex]\[
\frac{x^2 - 36}{x - 6} = x + 6 \quad \text{for} \; x \neq 6
\][/tex]
### Problem 2:
Calculate the limit [tex]\(\lim_{x \to 0} \frac{16x - 8x}{4x^2 + 2x}\)[/tex]
Solution:
First, simplify the expression inside the limit:
[tex]\[
\frac{16x - 8x}{4x^2 + 2x}
\][/tex]
Combine terms in the numerator:
[tex]\[
\frac{8x}{4x^2 + 2x}
\][/tex]
Factor out the common terms in both the numerator and the denominator. The numerator can be factored as:
[tex]\[
8x
\][/tex]
For the denominator:
[tex]\[
4x^2 + 2x = 2x(2x + 1)
\][/tex]
So, the expression becomes:
[tex]\[
\frac{8x}{2x(2x + 1)}
\][/tex]
We can cancel out the common factor [tex]\(2x\)[/tex]:
[tex]\[
\frac{8x}{2x(2x + 1)} = \frac{8}{2(2x + 1)} = \frac{8}{4x + 2}
\][/tex]
Now, take the limit as [tex]\(x\)[/tex] approaches 0:
[tex]\[
\lim_{x \to 0} \frac{8}{4x + 2}
\][/tex]
Substitute [tex]\(x = 0\)[/tex]:
[tex]\[
\frac{8}{4(0) + 2} = \frac{8}{2} = 4
\][/tex]
Therefore,
[tex]\(\lim_{x \to 0} \frac{16x - 8x}{4x^2 + 2x} = 4\)[/tex]
By following these detailed steps, we reach the simplified expressions and limits for the given mathematical problems.
