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What is the right-hand limit of the function [tex]f(x)=\frac{x^2+2x-3}{x-1}[/tex] as [tex]x[/tex] approaches 2?

A. 1
B. 2
C. 4
D. 5


Sagot :

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