lim _{x rightarrow 2}left[frac{1}{x-2}-frac{2 | vedclass

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Question

(C) We are given the limit: $\lim _{x \rightarrow 2}\left[\frac{1}{x-2}-\frac{2}{x^3-3 x^2+2 x}\right]$ First,factor the denominator of the second ter

Vedclass · Accepted Answer

$\frac{3}{2}$

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(C) We are given the limit: $\lim _{x \rightarrow 2}\left[\frac{1}{x-2}-\frac{2}{x^3-3 x^2+2 x}\right]$ First,factor the denominator of the second term: $x^3-3x^2+2x = x(x^2-3x+2) = x(x-2)(x-1)$ Now,substitute this back into the limit: $\lim _{x \rightarrow 2}\left[\frac{1}{x-2}-\frac{2}{x(x-2)(x-1)}\right]$ Find a common denominator: $\lim _{x \rightarrow 2}\left[\frac{x(x-1)-2}{x(x-2)(x-1)}\right]$ Simplify the numerator: $x^2-x-2 = (x-2)(x+1)$ Substitute the simplified numerator: $\lim _{x \rightarrow 2}\left[\frac{(x-2)(x+1)}{x(x-2)(x-1)}\right]$ Cancel the common factor $(x-2)$: $\lim _{x \rightarrow 2}\frac{x+1}{x(x-1)}$ Evaluate the limit by substituting $x=2$: $\frac{2+1}{2(2-1)} = \frac{3}{2(1)} = \frac{3}{2}$

$\lim _{x \rightarrow 2}\left[\frac{1}{x-2}-\frac{2}{x^3-3 x^2+2 x}\right]$ is equal to

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