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To determine the limit \(\lim_{x \to 1} \frac{1-2x+x^2}{1+\log x-x}\), let's walk through the solution step-by-step:
### Step 1: Analyze the Limit
First, both the numerator and the denominator functions need to be analyzed as \(x\) approaches 1.
#### Expression Analysis at \(x = 1\):
- For the numerator \(1 - 2x + x^2\):
[tex]\[ 1 - 2(1) + (1)^2 = 1 - 2 + 1 = 0 \][/tex]
- For the denominator \(1 + \log x - x\):
[tex]\[ 1 + \log(1) - 1 = 1 + 0 - 1 = 0 \][/tex]
Both the numerator and the denominator approach 0 as \(x\) approaches 1, creating an indeterminate form of \(\frac{0}{0}\).
### Step 2: Apply L'Hôpital's Rule
Since we have the indeterminate form \(\frac{0}{0}\), we can apply L'Hôpital's Rule. According to L'Hôpital's Rule, if \(\lim_{x \to c} \frac{f(x)}{g(x)}\) gives \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then:
[tex]\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \][/tex]
provided this latter limit exists.
#### Differentiating the Numerator and Denominator:
- Numerator \(1 - 2x + x^2\):
[tex]\[ \frac{d}{dx}\left(1 - 2x + x^2\right) = -2 + 2x \][/tex]
- Denominator \(1 + \log x - x\):
[tex]\[ \frac{d}{dx}\left(1 + \log x - x\right) = \frac{1}{x} - 1 \][/tex]
### Step 3: Apply the Derivatives:
Now, apply the derivatives and find the limit:
[tex]\[ \lim_{x \to 1} \frac{-2 + 2x}{\frac{1}{x} - 1} \][/tex]
Substitute \(x = 1\) into the derivatives:
- For the numerator \(-2 + 2(1) = -2 + 2 = 0\)
- For the denominator \(\frac{1}{1} - 1 = 1 - 1 = 0\)
Once again, we have the form \(\frac{0}{0}\), so we apply L'Hôpital's Rule again by differentiating the numerator and the denominator a second time.
- Second derivative of numerator \(-2 + 2x\):
[tex]\[ \frac{d}{dx}\left(-2 + 2x\right) = 2 \][/tex]
- Second derivative of denominator \(\frac{1}{x} - 1\):
[tex]\[ \frac{d}{dx}\left(\frac{1}{x} - 1\right) = -\frac{1}{x^2} \][/tex]
### Step 4: Apply the Second Derivatives:
[tex]\[ \lim_{x \to 1} \frac{2}{-\frac{1}{x^2}} \][/tex]
Substitute \(x = 1\):
[tex]\[ \frac{2}{-\frac{1}{1^2}} = \frac{2}{-1} = -2 \][/tex]
Thus, the limit is:
[tex]\[ \boxed{-2} \][/tex]
### Step 1: Analyze the Limit
First, both the numerator and the denominator functions need to be analyzed as \(x\) approaches 1.
#### Expression Analysis at \(x = 1\):
- For the numerator \(1 - 2x + x^2\):
[tex]\[ 1 - 2(1) + (1)^2 = 1 - 2 + 1 = 0 \][/tex]
- For the denominator \(1 + \log x - x\):
[tex]\[ 1 + \log(1) - 1 = 1 + 0 - 1 = 0 \][/tex]
Both the numerator and the denominator approach 0 as \(x\) approaches 1, creating an indeterminate form of \(\frac{0}{0}\).
### Step 2: Apply L'Hôpital's Rule
Since we have the indeterminate form \(\frac{0}{0}\), we can apply L'Hôpital's Rule. According to L'Hôpital's Rule, if \(\lim_{x \to c} \frac{f(x)}{g(x)}\) gives \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then:
[tex]\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \][/tex]
provided this latter limit exists.
#### Differentiating the Numerator and Denominator:
- Numerator \(1 - 2x + x^2\):
[tex]\[ \frac{d}{dx}\left(1 - 2x + x^2\right) = -2 + 2x \][/tex]
- Denominator \(1 + \log x - x\):
[tex]\[ \frac{d}{dx}\left(1 + \log x - x\right) = \frac{1}{x} - 1 \][/tex]
### Step 3: Apply the Derivatives:
Now, apply the derivatives and find the limit:
[tex]\[ \lim_{x \to 1} \frac{-2 + 2x}{\frac{1}{x} - 1} \][/tex]
Substitute \(x = 1\) into the derivatives:
- For the numerator \(-2 + 2(1) = -2 + 2 = 0\)
- For the denominator \(\frac{1}{1} - 1 = 1 - 1 = 0\)
Once again, we have the form \(\frac{0}{0}\), so we apply L'Hôpital's Rule again by differentiating the numerator and the denominator a second time.
- Second derivative of numerator \(-2 + 2x\):
[tex]\[ \frac{d}{dx}\left(-2 + 2x\right) = 2 \][/tex]
- Second derivative of denominator \(\frac{1}{x} - 1\):
[tex]\[ \frac{d}{dx}\left(\frac{1}{x} - 1\right) = -\frac{1}{x^2} \][/tex]
### Step 4: Apply the Second Derivatives:
[tex]\[ \lim_{x \to 1} \frac{2}{-\frac{1}{x^2}} \][/tex]
Substitute \(x = 1\):
[tex]\[ \frac{2}{-\frac{1}{1^2}} = \frac{2}{-1} = -2 \][/tex]
Thus, the limit is:
[tex]\[ \boxed{-2} \][/tex]
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