Let's find the limit of the expression [tex]\(\frac{x^2 + 3x - 4}{x - 1}\)[/tex] as [tex]\(x\)[/tex] approaches 1.
First, we notice that directly substituting [tex]\(x = 1\)[/tex] into the expression would give us a division by zero, which means we have to take another approach to solve this limit.
Let's start by factoring the numerator [tex]\(x^2 + 3x - 4\)[/tex]:
[tex]\[ x^2 + 3x - 4 \][/tex]
We can factor this quadratic expression:
[tex]\[ x^2 + 3x - 4 = (x + 4)(x - 1) \][/tex]
So, the given expression becomes:
[tex]\[ \frac{(x + 4)(x - 1)}{x - 1} \][/tex]
Now, we can cancel the common term [tex]\((x - 1)\)[/tex] in both the numerator and the denominator:
[tex]\[ \frac{(x + 4)\cancel{(x - 1)}}{\cancel{(x - 1)}} = x + 4 \quad \text{for } x \neq 1 \][/tex]
Now we need to find the limit of [tex]\(x + 4\)[/tex] as [tex]\(x\)[/tex] approaches 1:
[tex]\[ \lim_{x \to 1} (x + 4) \][/tex]
As [tex]\(x\)[/tex] approaches 1, [tex]\(x + 4\)[/tex] approaches:
[tex]\[ 1 + 4 = 5 \][/tex]
So, the limit is:
[tex]\[ \lim_{x \to 1} \frac{x^2 + 3x - 4}{x - 1} = 5 \][/tex]
