Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 4

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PROBLEM:

Evaluate lim⁡x→2(x3−x2−x−22x3−5x2+5x−6) \displaystyle \lim\limits_{x\to 2}\left(\frac{x^3-x^2-x-2}{2x^3-5x^2+5x-6}\right).

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 SOLUTION:

A straight substitution of x=2x=2 leads to the indeterminate form 00\frac{0}{0} which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

lim⁡x→2(x3−x2−x−22x3−5x2+5x−6)=lim⁡x→2((x−2)(x2+x+1)(x−2)(2x2−x+3))=lim⁡x→2(x2+x+12x2−x+3)=22+2+12(2)2−2+3=4+2+18−2+3=79  (Answer)\begin{align*} \lim\limits_{x\to 2}\left(\frac{x^3-x^2-x-2}{2x^3-5x^2+5x-6}\right)&=\lim\limits_{x\to 2}\left(\frac{\left(x-2\right)\left(x^2+x+1\right)}{\left(x-2\right)\left(2x^2-x+3\right)}\right)\\ \\ & =\lim\limits_{x\to 2}\left(\frac{x^2+x+1}{2x^2-x+3}\right)\\ \\ & =\frac{2^2+2+1}{2\left(2\right)^2-2+3}\\ \\ & =\frac{4+2+1}{8-2+3}\\ \\ & =\frac{7}{9} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \\ \end{align*}
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