To find the limit [tex]\(\lim_ {x \rightarrow 1} \frac{x^2-x \log_e x + \log_e x - 1}{x-1}\)[/tex], we should carefully examine the expression. Directly substituting [tex]\(x = 1\)[/tex] into the function results in an indeterminate form [tex]\(\frac{0}{0}\)[/tex]. Thus, we'll apply L'Hôpital's rule, which states that for limits of the form [tex]\(\frac{0}{0}\)[/tex] or [tex]\(\frac{\infty}{\infty}\)[/tex], we can take the derivative of the numerator and the denominator separately and then find the limit.
First, let's rewrite the limit:
[tex]\[
\lim_{x \to 1} \frac{x^2 - x \log_e x + \log_e x - 1}{x - 1}
\][/tex]
To apply L'Hôpital's rule, we need to take the derivative of the numerator and the denominator with respect to [tex]\(x\)[/tex].
1. Derivative of the Numerator:
- The numerator is [tex]\(f(x) = x^2 - x \log_e x + \log_e x - 1\)[/tex].
- The derivative [tex]\(f'(x)\)[/tex] is found as follows:
[tex]\[
\frac{d}{dx} \left( x^2 - x \log_e x + \log_e x - 1 \right)
\][/tex]
- Derivative of [tex]\(x^2\)[/tex] is [tex]\(2x\)[/tex].
- Derivative of [tex]\(-x \log_e x\)[/tex] is calculated using the product rule:
[tex]\[
\frac{d}{dx} (-x \log_e x) = -\left(\log_e x + x \cdot \frac{1}{x}\right) = -\log_e x - 1
\][/tex]
- Derivative of [tex]\(\log_e x\)[/tex] is [tex]\(\frac{1}{x}\)[/tex].
- Derivative of [tex]\(-1\)[/tex] is [tex]\(0\)[/tex].
Thus, combining all these together:
[tex]\[
f'(x) = 2x - (\log_e x + 1) + \frac{1}{x} = 2x - \log_e x - 1 + \frac{1}{x}
\][/tex]
2. Derivative of the Denominator:
- The denominator is [tex]\(g(x) = x - 1\)[/tex].
- The derivative [tex]\(g'(x)\)[/tex] is:
[tex]\[
\frac{d}{dx} (x - 1) = 1
\][/tex]
Now, by L'Hôpital's rule, we take the limit of the ratio of the derivatives:
[tex]\[
\lim_{x \to 1} \frac{f'(x)}{g'(x)} = \lim_{x \to 1} \left(2x - \log_e x - 1 + \frac{1}{x}\right)
\][/tex]
Substitute [tex]\(x = 1\)[/tex] into the simplified expression:
[tex]\[
2(1) - \log_e (1) - 1 + \frac{1}{1}
\][/tex]
Simplify the expression:
- [tex]\(\log_e (1) = 0\)[/tex]
- [tex]\(2(1) = 2\)[/tex]
- [tex]\(\frac{1}{1} = 1\)[/tex]
Thus:
[tex]\[
2 - 0 - 1 + 1 = 2
\][/tex]
Therefore, the limit is:
[tex]\[
\lim_{x \rightarrow 1} \frac{x^2 - x \log_e x + \log_e x - 1}{x - 1} = 2
\][/tex]
