To solve [tex]\(\lim_{x \rightarrow 1} \frac{x^5 + x^3 - 2}{x^3 - 1}\)[/tex], we will proceed with a step-by-step approach.
1. Evaluate the function at [tex]\(x = 1\)[/tex]:
Let's first plug [tex]\(x = 1\)[/tex] into the function to determine whether we get an indeterminate form.
[tex]\[
\frac{1^5 + 1^3 - 2}{1^3 - 1} = \frac{1 + 1 - 2}{1 - 1} = \frac{0}{0}
\][/tex]
This gives us an indeterminate form [tex]\( \frac{0}{0} \)[/tex]. Therefore, we should simplify the expression or use other techniques such as L'Hôpital's Rule.
2. Simplify the expression by factoring:
Notice that the denominator [tex]\(x^3 - 1\)[/tex] can be factored:
[tex]\[
x^3 - 1 = (x - 1)(x^2 + x + 1)
\][/tex]
Now we factor the numerator [tex]\(x^5 + x^3 - 2\)[/tex] similarly. However, factoring the numerator directly is not straightforward. Instead, let's apply L'Hôpital's Rule because we have [tex]\(\frac{0}{0}\)[/tex].
3. Apply L'Hôpital's Rule:
L'Hôpital's Rule states that if we have an indeterminate form of [tex]\(\frac{0}{0}\)[/tex] or [tex]\(\frac{\infty}{\infty}\)[/tex], then:
[tex]\[
\lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \lim_{x \rightarrow c} \frac{f'(x)}{g'(x)}
\][/tex]
provided the limit on the right exists. Here, [tex]\( f(x) = x^5 + x^3 - 2 \)[/tex] and [tex]\( g(x) = x^3 - 1 \)[/tex].
First, compute the derivatives of the numerator and the denominator.
[tex]\[
f'(x) = \frac{d}{dx}(x^5 + x^3 - 2) = 5x^4 + 3x^2
\][/tex]
[tex]\[
g'(x) = \frac{d}{dx}(x^3 - 1) = 3x^2
\][/tex]
4. Substitute into L'Hôpital's Rule:
Applying L'Hôpital's Rule, we get:
[tex]\[
\lim_{x \rightarrow 1} \frac{x^5 + x^3 - 2}{x^3 - 1} = \lim_{x \rightarrow 1} \frac{5x^4 + 3x^2}{3x^2}
\][/tex]
5. Simplify the new function:
Simplify the fraction inside the limit:
[tex]\[
\frac{5x^4 + 3x^2}{3x^2} = \frac{5x^4}{3x^2} + \frac{3x^2}{3x^2} = \frac{5x^2}{3} + 1
\][/tex]
6. Evaluate the limit:
Now, substitute [tex]\(x = 1\)[/tex] into the simplified function:
[tex]\[
\lim_{x \rightarrow 1} \left( \frac{5x^2}{3} + 1 \right) = \frac{5(1)^2}{3} + 1 = \frac{5}{3} + 1 = \frac{5}{3} + \frac{3}{3} = \frac{8}{3}
\][/tex]
Therefore, the limit is:
[tex]\[
\boxed{\frac{8}{3}}
\][/tex]
