Limit of a Function in Indeterminate Form| Differential and Integral Calculus| Feliciano and Uy| Exercise 1.3, Problem 4|

Evaluate

 \lim\limits_{x\to 2}\left(\frac{x^3-x^2-x-2}{2x^3-5x^2+5x-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^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}

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