npenes lim _(xarrow 1)(x^3-x)/(x^2)-2x+1

To evaluate the limit \(\lim_{x \to 1} \frac{x^3 - x}{x^2 - 2x + 1}\), we first simplify the expression.<br /><br />Notice that the denominator can be factored:<br />\[ x^2 - 2x + 1 = (x - 1)^2 \]<br /><br />Next, we factor the numerator:<br />\[ x^3 - x = x(x^2 - 1) = x(x - 1)(x + 1) \]<br /><br />So the limit becomes:<br />\[ \lim_{x \to 1} \frac{x(x - 1)(x + 1)}{(x - 1)^2} \]<br /><br />We can cancel the common factor \((x - 1)\) in the numerator and denominator:<br />\[ \lim_{x \to 1} \frac{x(x + 1)}{x - 1} = \lim_{x \to 1} \frac{x(x + 1)}{x - 1} \]<br /><br />Now, substitute \(x = 1\) into the simplified expression:<br />\[ \frac{1(1 + 1)}{1 - 1} = \frac{2}{0} \]<br /><br />Since this results in an indeterminate form \(\frac{2}{0}\), we need to re-evaluate our simplification. Let's instead use L'Hôpital's Rule, which is applicable for limits of the form \(\frac{0}{0}\).<br /><br />Differentiate the numerator and the denominator:<br />\[ \text{Numerator: } \frac{d}{dx}(x^3 - x) = 3x^2 - 1 \]<br />\[ \text{Denominator: } \frac{d}{dx}(x^2 - 2x + 1) = 2x - 2 \]<br /><br />Now apply L'Hôpital's Rule:<br />\[ \lim_{x \to 1} \frac{x^3 - x}{x^2 - 2x + 1} = \lim_{x \to 1} \frac{3x^2 - 1}{2x - 2} \]<br /><br />Substitute \(x = 1\) into the new expression:<br />\[ \frac{3(1)^2 - 1}{2(1) - 2} = \frac{3 - 1}{2 - 2} = \frac{2}{0} \]<br /><br />This still results in an indeterminate form. Let's try another approach by factoring and simplifying directly.<br /><br />Factor the numerator:<br />\[ x^3 - x = x(x^2 - 1) = x(x - 1)(x + 1) \]<br /><br />Factor the denominator:<br />\[ x^2 - 2x + 1 = (x - 1)^2 \]<br /><br />Cancel the common factor \((x - 1)\):<br />\[ \frac{x(x - 1)(x + 1)}{(x - 1)^2} = \frac{x(x + 1)}{x - 1} \]<br /><br />Now, substitute \(x = 1\):<br />\[ \frac{1(1 + 1)}{1 - 1} = \frac{2}{0} \]<br /><br />Since this still results in an indeterminate form, we conclude that the limit does not exist in a finite sense. Therefore, the limit is undefined.