Limit of a Function in Indeterminate Form| Differential and Integral Calculus| Feliciano and Uy| Exercise 1.3, Problem 2|

Evaluate

\lim\limits_{x\to 2}\left(\frac{x^2+2x-8}{3x-6}\right)

SOLUTION:

A straight substitution of x=2 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 2}\left(\frac{x^2+2x-8}{3x-6}\right)=\lim\limits_{x\to 2}\left(\frac{\left(x+4\right)\left(x-2\right)}{3\left(x-2\right)}\right)

=\lim\limits_{x\to 2}\left(\frac{x+4}{3}\right)

=\frac{2+4}{3}

=\frac{6}{3}

=2

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