To solve for the right-hand limit of the function [tex]\( f(x) = \frac{x^2 + 2x - 3}{x - 1} \)[/tex] as [tex]\( x \)[/tex] approaches 2, let's go through the steps methodically.
1. First, understand the function:
The function given is [tex]\( f(x) = \frac{x^2 + 2x - 3}{x - 1} \)[/tex].
2. Simplify the function if possible:
We notice that the numerator [tex]\((x^2 + 2x - 3)\)[/tex] can be factored. Let's factor it:
[tex]\[
x^2 + 2x - 3 = (x + 3)(x - 1)
\][/tex]
So the function becomes:
[tex]\[
f(x) = \frac{(x + 3)(x - 1)}{x - 1}
\][/tex]
For [tex]\( x \neq 1 \)[/tex], we can cancel the [tex]\((x - 1)\)[/tex] terms in the numerator and the denominator:
[tex]\[
f(x) = x + 3 \quad \text{for} \quad x \neq 1
\][/tex]
3. Determine the limit:
Now, we need to find the right-hand limit as [tex]\( x \)[/tex] approaches 2 of the simplified function [tex]\( f(x) = x + 3 \)[/tex].
[tex]\[
\lim_{{x \to 2^+}} (x + 3)
\][/tex]
4. Evaluate the limit:
Plug in [tex]\( x = 2 \)[/tex] into the simplified function:
[tex]\[
\lim_{{x \to 2^+}} (x + 3) = 2 + 3 = 5
\][/tex]
Thus, the right-hand limit of the function [tex]\( f(x) = \frac{x^2 + 2x - 3}{x - 1} \)[/tex] as [tex]\( x \)[/tex] approaches 2 is [tex]\( 5 \)[/tex].
So, the correct answer is:
D. 5
