To solve the limit [tex]\(\lim_ {x \rightarrow 3} \frac{x^5 - 3^5}{x - 3}\)[/tex], let's follow these steps:
1. Recognize the indeterminate form: When [tex]\( x \)[/tex] approaches 3, both the numerator and the denominator approach 0, leading to an indeterminate form of [tex]\(\frac{0}{0}\)[/tex].
2. Factor the numerator: Notice that the numerator is a difference of powers, [tex]\( x^5 - 3^5 \)[/tex]. This can be factored using the formula for the difference of powers:
[tex]\[
a^5 - b^5 = (a - b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4)
\][/tex]
In this case, [tex]\( a = x \)[/tex] and [tex]\( b = 3 \)[/tex], so:
[tex]\[
x^5 - 3^5 = (x - 3)(x^4 + 3x^3 + 9x^2 + 27x + 81)
\][/tex]
3. Simplify the expression: Substitute the factorization back into the limit:
[tex]\[
\frac{x^5 - 3^5}{x - 3} = \frac{(x - 3)(x^4 + 3x^3 + 9x^2 + 27x + 81)}{x - 3}
\][/tex]
Cancel the common factor [tex]\((x - 3)\)[/tex] in the numerator and denominator:
[tex]\[
\frac{(x - 3)(x^4 + 3x^3 + 9x^2 + 27x + 81)}{x - 3} = x^4 + 3x^3 + 9x^2 + 27x + 81
\][/tex]
4. Evaluate the limit: Now that we have simplified the expression, we can directly substitute [tex]\( x = 3 \)[/tex]:
[tex]\[
\lim_{x \to 3} (x^4 + 3x^3 + 9x^2 + 27x + 81)
\][/tex]
Replace [tex]\( x \)[/tex] with 3:
[tex]\[
3^4 + 3 \cdot 3^3 + 9 \cdot 3^2 + 27 \cdot 3 + 81
\][/tex]
Calculate each term separately:
[tex]\[
3^4 = 81, \quad 3 \cdot 3^3 = 3 \cdot 27 = 81, \quad 9 \cdot 3^2 = 9 \cdot 9 = 81, \quad 27 \cdot 3 = 81, \quad 81
\][/tex]
5. Sum the terms: Add the results together:
[tex]\[
81 + 81 + 81 + 81 + 81 = 405
\][/tex]
Therefore, the limit is:
[tex]\[
\boxed{405}
\][/tex]
