Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 3

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PROBLEM:

Evaluate lim⁡x→3(x3−13x+12x3−14x+15) \displaystyle \lim\limits_{x\to 3}\left(\frac{x^3-13x+12}{x^3-14x+15}\right).

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SOLUTION:

A straight substitution of x=3x=3 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→3(x3−13x+12x3−14x+15)=lim⁡x→3((x−3)(x2+3x−4)(x−3)(x2+3x−5))=lim⁡x→3(x2+3x−4x2+3x−5)=(3)2+3(3)−4(3)2+3(3)−5=9+9−49+9−5=1413  (Answer)\begin{align*} \lim\limits_{x\to 3}\left(\frac{x^3-13x+12}{x^3-14x+15}\right)& =\lim\limits_{x\to 3}\left(\frac{\left(x-3\right)\left(x^2+3x-4\right)}{\left(x-3\right)\left(x^2+3x-5\right)}\right)\\ \\ & =\lim\limits_{x\to 3}\left(\frac{x^2+3x-4}{x^2+3x-5}\right)\\ \\ &=\frac{\left(3\right)^2+3\left(3\right)-4}{\left(3\right)^2+3\left(3\right)-5}\\ \\ & =\frac{9+9-4}{9+9-5}\\ \\ & =\frac{14}{13} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \\ \end{align*}
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