Certainly! Let's analyze the given function [tex]\( f(x) = \frac{x^2 + 3x - 4}{x - 1} \)[/tex] and find the limit as [tex]\( x \)[/tex] approaches 1. Here’s a detailed, step-by-step solution:
1. Function Analysis:
We start with the function [tex]\( f(x) = \frac{x^2 + 3x - 4}{x - 1} \)[/tex].
2. Factor the Numerator:
We recognize that the numerator [tex]\( x^2 + 3x - 4 \)[/tex] can be factored. The factors of [tex]\(-4\)[/tex] that sum up to [tex]\(3\)[/tex] are [tex]\(4\)[/tex] and [tex]\(-1\)[/tex]. Hence, we can rewrite the numerator:
[tex]\[
x^2 + 3x - 4 = (x + 4)(x - 1)
\][/tex]
3. Simplify the Function:
After factoring, the function is:
[tex]\[
f(x) = \frac{(x + 4)(x - 1)}{x - 1}
\][/tex]
We notice that [tex]\((x - 1)\)[/tex] in the numerator and denominator can cancel each other out, but only if [tex]\( x \neq 1 \)[/tex]. Thus, the simplified form of the function (for [tex]\( x \neq 1 \)[/tex]) is:
[tex]\[
f(x) = x + 4
\][/tex]
4. Finding the Limit:
To find the limit as [tex]\( x \)[/tex] approaches 1, we use the simplified form of the function:
[tex]\[
\lim_{x \to 1} f(x) = \lim_{x \to 1} (x + 4)
\][/tex]
Substituting [tex]\( x = 1 \)[/tex] into [tex]\( x + 4 \)[/tex], we get:
[tex]\[
f(1) = 1 + 4 = 5
\][/tex]
Therefore, the limit as [tex]\( x \)[/tex] approaches 1 for the function [tex]\( \frac{x^2 + 3x - 4}{x - 1} \)[/tex] is [tex]\( 5 \)[/tex].
