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To evaluate the limit [tex]\(\lim_{x \rightarrow 1}\left[\frac{x-2}{x^2-1}-\frac{1}{x^3-3 x^2+2 x}\right]\)[/tex], we will go through the following steps:
1. Simplify the individual expressions:
First term:
[tex]\[ \frac{x-2}{x^2-1} \][/tex]
Notice that [tex]\(x^2 - 1\)[/tex] can be factored as:
[tex]\[ x^2 - 1 = (x + 1)(x - 1) \][/tex]
Hence, the expression becomes:
[tex]\[ \frac{x-2}{(x+1)(x-1)} \][/tex]
Second term:
[tex]\[ \frac{1}{x^3 - 3x^2 + 2x} \][/tex]
The denominator [tex]\(x^3 - 3x^2 + 2x\)[/tex] can be factored by factoring out an [tex]\(x\)[/tex] first:
[tex]\[ x^3 - 3x^2 + 2x = x(x^2 - 3x + 2) \][/tex]
Further factoring [tex]\(x^2 - 3x + 2\)[/tex] gives:
[tex]\[ x^2 - 3x + 2 = (x - 1)(x - 2) \][/tex]
Therefore, the expression can be written as:
[tex]\[ \frac{1}{x(x-1)(x-2)} \][/tex]
2. Combine the expressions:
We need to combine the two terms into a single expression:
[tex]\[ \frac{x-2}{(x+1)(x-1)} - \frac{1}{x(x-1)(x-2)} \][/tex]
The common denominator for these fractions is [tex]\((x+1)(x-1)x(x-2)\)[/tex], so each fraction needs to be rewritten with this common denominator:
For the first term:
[tex]\[ \frac{x-2}{(x+1)(x-1)} = \frac{(x-2)x}{(x+1)x(x-1)(x-2)} = \frac{x(x-2)}{(x+1)x(x-1)(x-2)} \][/tex]
For the second term:
[tex]\[ \frac{1}{x(x-1)(x-2)} = \frac{x+1}{(x+1)x(x-1)(x-2)} \][/tex]
Now subtract the second fraction from the first:
[tex]\[ \frac{x(x-2) - (x+1)}{(x+1)x(x-1)(x-2)} \][/tex]
3. Simplify the numerator:
Simplify the expression in the numerator:
[tex]\[ x(x-2) - (x+1) = x^2 - 2x - x - 1 = x^2 - 3x - 1 \][/tex]
Thus, our combined expression now looks like:
[tex]\[ \frac{x^2 - 3x - 1}{(x+1)x(x-1)(x-2)} \][/tex]
4. Evaluate the limit as [tex]\(x \rightarrow 1\)[/tex]:
We now need to evaluate:
[tex]\[ \lim_{x \to 1} \frac{x^2 - 3x - 1}{(x+1)x(x-1)(x-2)} \][/tex]
At [tex]\(x = 1\)[/tex], the denominator becomes zero since it contains the term [tex]\((x-1)\)[/tex]:
[tex]\[ (1+1)(1)(1-1)(1-2) = 0 \][/tex]
Hence, this makes the entire expression undefined because we're essentially dividing by zero. This indicates that the limit does not exist in the usual sense and results in an infinite discontinuity.
Therefore, the limit is:
[tex]\[ \lim_{x \rightarrow 1}\left[\frac{x-2}{x^2-1}-\frac{1}{x^3-3 x^2+2 x}\right] = \infty \][/tex]
Additionally, the simplified form of the expression before taking the limit is:
[tex]\[ \frac{x^2 - 3x - 1}{(x^3 - 2x^2 - x + 2)} \][/tex]
Thus, the final answer:
[tex]\[ \lim_{x \to 1} \left[\frac{x-2}{x^2-1} - \frac{1}{x^3 - 3x^2 + 2x}\right] = \infty \][/tex]
And the simplified expression:
[tex]\[ \frac{x^2 - 3x - 1}{(x^3 - 2x^2 - x + 2)} \][/tex]
1. Simplify the individual expressions:
First term:
[tex]\[ \frac{x-2}{x^2-1} \][/tex]
Notice that [tex]\(x^2 - 1\)[/tex] can be factored as:
[tex]\[ x^2 - 1 = (x + 1)(x - 1) \][/tex]
Hence, the expression becomes:
[tex]\[ \frac{x-2}{(x+1)(x-1)} \][/tex]
Second term:
[tex]\[ \frac{1}{x^3 - 3x^2 + 2x} \][/tex]
The denominator [tex]\(x^3 - 3x^2 + 2x\)[/tex] can be factored by factoring out an [tex]\(x\)[/tex] first:
[tex]\[ x^3 - 3x^2 + 2x = x(x^2 - 3x + 2) \][/tex]
Further factoring [tex]\(x^2 - 3x + 2\)[/tex] gives:
[tex]\[ x^2 - 3x + 2 = (x - 1)(x - 2) \][/tex]
Therefore, the expression can be written as:
[tex]\[ \frac{1}{x(x-1)(x-2)} \][/tex]
2. Combine the expressions:
We need to combine the two terms into a single expression:
[tex]\[ \frac{x-2}{(x+1)(x-1)} - \frac{1}{x(x-1)(x-2)} \][/tex]
The common denominator for these fractions is [tex]\((x+1)(x-1)x(x-2)\)[/tex], so each fraction needs to be rewritten with this common denominator:
For the first term:
[tex]\[ \frac{x-2}{(x+1)(x-1)} = \frac{(x-2)x}{(x+1)x(x-1)(x-2)} = \frac{x(x-2)}{(x+1)x(x-1)(x-2)} \][/tex]
For the second term:
[tex]\[ \frac{1}{x(x-1)(x-2)} = \frac{x+1}{(x+1)x(x-1)(x-2)} \][/tex]
Now subtract the second fraction from the first:
[tex]\[ \frac{x(x-2) - (x+1)}{(x+1)x(x-1)(x-2)} \][/tex]
3. Simplify the numerator:
Simplify the expression in the numerator:
[tex]\[ x(x-2) - (x+1) = x^2 - 2x - x - 1 = x^2 - 3x - 1 \][/tex]
Thus, our combined expression now looks like:
[tex]\[ \frac{x^2 - 3x - 1}{(x+1)x(x-1)(x-2)} \][/tex]
4. Evaluate the limit as [tex]\(x \rightarrow 1\)[/tex]:
We now need to evaluate:
[tex]\[ \lim_{x \to 1} \frac{x^2 - 3x - 1}{(x+1)x(x-1)(x-2)} \][/tex]
At [tex]\(x = 1\)[/tex], the denominator becomes zero since it contains the term [tex]\((x-1)\)[/tex]:
[tex]\[ (1+1)(1)(1-1)(1-2) = 0 \][/tex]
Hence, this makes the entire expression undefined because we're essentially dividing by zero. This indicates that the limit does not exist in the usual sense and results in an infinite discontinuity.
Therefore, the limit is:
[tex]\[ \lim_{x \rightarrow 1}\left[\frac{x-2}{x^2-1}-\frac{1}{x^3-3 x^2+2 x}\right] = \infty \][/tex]
Additionally, the simplified form of the expression before taking the limit is:
[tex]\[ \frac{x^2 - 3x - 1}{(x^3 - 2x^2 - x + 2)} \][/tex]
Thus, the final answer:
[tex]\[ \lim_{x \to 1} \left[\frac{x-2}{x^2-1} - \frac{1}{x^3 - 3x^2 + 2x}\right] = \infty \][/tex]
And the simplified expression:
[tex]\[ \frac{x^2 - 3x - 1}{(x^3 - 2x^2 - x + 2)} \][/tex]
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