Limit of a Function in Indeterminate Form| Differential and Integral Calculus| Feliciano and Uy| Exercise 1.3, Problem 3|

Evaluate

 \lim\limits_{x\to 3}\left(\frac{x^3-13x+12}{x^3-14x+15}\right)

SOLUTION:

A straight substitution of x=3 leads to the indeterminate form  \frac{0}{0}  which is meaningless.

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

\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}

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