To compute the limit [tex]\(\lim_{x \rightarrow 1} \frac{\frac{1}{x+2} - \frac{1}{3}}{x-1}\)[/tex], we start by analyzing the given expression step-by-step.
First, consider the function:
[tex]\[ f(x) = \frac{\frac{1}{x+2} - \frac{1}{3}}{x-1} \][/tex]
We need to find [tex]\(\lim_{x \rightarrow 1} f(x)\)[/tex].
The expression inside the limit can be simplified. Combine the terms in the numerator:
[tex]\[ \frac{1}{x+2} - \frac{1}{3} = \frac{3 - (x+2)}{3(x+2)} = \frac{3 - x - 2}{3(x+2)} = \frac{1 - x}{3(x+2)} \][/tex]
Thus, the function becomes:
[tex]\[ f(x) = \frac{\frac{1 - x}{3(x+2)}}{x-1} \][/tex]
This simplifies to:
[tex]\[ f(x) = \frac{1 - x}{3(x+2)(x - 1)} \][/tex]
Notice that the numerator [tex]\(1 - x\)[/tex] is [tex]\(-(x - 1)\)[/tex]. Substitute this in:
[tex]\[ f(x) = \frac{-(x - 1)}{3(x+2)(x - 1)} \][/tex]
Cancel the common term [tex]\((x - 1)\)[/tex] in the numerator and the denominator:
[tex]\[ f(x) = \frac{-1}{3(x + 2)} \][/tex]
Now, find the limit as [tex]\(x\)[/tex] approaches 1:
[tex]\[ \lim_{x \rightarrow 1} f(x) = \lim_{x \rightarrow 1} \frac{-1}{3(x + 2)} \][/tex]
Substitute [tex]\(x = 1\)[/tex]:
[tex]\[ f(1) = \frac{-1}{3(1 + 2)} = \frac{-1}{3 \cdot 3} = \frac{-1}{9} \][/tex]
The limit is:
[tex]\[ \lim_{x \rightarrow 1} \frac{\frac{1}{x+2} - \frac{1}{3}}{x-1} = \frac{-1}{9} \][/tex]
This result is [tex]\(-\frac{1}{9}\)[/tex]. Converting this fraction to a decimal:
[tex]\[ -\frac{1}{9} \approx -0.1111 \][/tex]
Thus, the best answer from the given choices is:
a. [tex]\(\lim_{x \rightarrow 1} f(x) = -0.1111\)[/tex]
So, the correct answer is:
a. [tex]\(\lim_{x \rightarrow 1} f(x) = -0.1111\)[/tex]
