To evaluate the limit
[tex]\[
\lim_{x \rightarrow 0} \frac{(x+1)^8 - 1}{(x+1)^2 - 1},
\][/tex]
we start by considering the function:
[tex]\[
\frac{(x+1)^8 - 1}{(x+1)^2 - 1}.
\][/tex]
Substituting [tex]\( x = 0 \)[/tex] directly gives:
[tex]\[
\frac{(0+1)^8 - 1}{(0+1)^2 - 1} = \frac{1 - 1}{1 - 1} = \frac{0}{0},
\][/tex]
which is an indeterminate form. Therefore, we must simplify the expression.
First, let's factor the numerator and the denominator:
The numerator can be rewritten as:
[tex]\[
(x+1)^8 - 1^8
\][/tex]
Using the difference of powers formula, [tex]\(a^8 - b^8 = (a - b)(a^7 + a^6b + a^5b^2 + a^4b^3 + a^3b^4 + a^2b^5 + ab^6 + b^7)\)[/tex], with [tex]\(a = x+1\)[/tex] and [tex]\(b = 1\)[/tex], we get:
[tex]\[
(x+1)^8 - 1 = (x+1 - 1)\left((x+1)^7 + (x+1)^6(1) + (x+1)^5(1^2) + (x+1)^4(1^3) + (x+1)^3(1^4) + (x+1)^2(1^5) + (x+1)(1^6) + 1^7\right).
\][/tex]
This simplifies to:
[tex]\[
(x+1 - 1)\left((x+1)^7 + (x+1)^6 + (x+1)^5 + (x+1)^4 + (x+1)^3 + (x+1)^2 + (x+1) + 1\right) = x \left((x+1)^7 + (x+1)^6 + (x+1)^5 + (x+1)^4 + (x+1)^3 + (x+1)^2 + (x+1) + 1\right).
\][/tex]
The denominator is:
[tex]\[
(x+1)^2 - 1^2
\][/tex]
Using the difference of squares formula, [tex]\(a^2 - b^2 = (a - b)(a + b)\)[/tex], with [tex]\(a = x+1\)[/tex] and [tex]\(b = 1\)[/tex], we get:
[tex]\[
(x+1)^2 - 1 = (x+1 - 1)(x+1 + 1) = x (x+2).
\][/tex]
Therefore, our limit expression becomes:
[tex]\[
\frac{x \left((x+1)^7 + (x+1)^6 + (x+1)^5 + (x+1)^4 + (x+1)^3 + (x+1)^2 + (x+1) + 1\right)}{x (x+2)}.
\][/tex]
As long as [tex]\(x \neq 0\)[/tex], we can cancel [tex]\(x\)[/tex] from numerator and denominator:
[tex]\[
\frac{\left((x+1)^7 + (x+1)^6 + (x+1)^5 + (x+1)^4 + (x+1)^3 + (x+1)^2 + (x+1) + 1\right)}{(x+2)}.
\][/tex]
Now we take the limit as [tex]\(x\)[/tex] approaches 0:
[tex]\[
\lim_{x \rightarrow 0} \frac{\left((x+1)^7 + (x+1)^6 + (x+1)^5 + (x+1)^4 + (x+1)^3 + (x+1)^2 + (x+1) + 1\right)}{(x+2)}.
\][/tex]
Substituting [tex]\( x = 0 \)[/tex]:
[tex]\[
\frac{(1)^7 + (1)^6 + (1)^5 + (1)^4 + (1)^3 + (1)^2 + (1) + 1}{0 + 2} = \frac{1 + 1 + 1 + 1 + 1 + 1 + 1 + 1}{2} = \frac{8}{2} = 4.
\][/tex]
Hence, the limit is:
[tex]\[
4.
\][/tex]
