# 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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