Differential and Integral Calculus by Feliciano and Uy, Exercise 1.2, Problem 8

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PROBLEM:

Evaluate lim⁡x→0(3x+2x2−2x+4)\displaystyle \lim\limits_{x\to 0}\left(\frac{3x+2}{x^2-2x+4}\right).

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SOLUTION:

Plug in the value x=0.

lim⁡x→0(3x+2x2−2x+4)=3(0)+2(0)2−2(0)+4=0+20−0+4=24=12  (Answer)\begin{align*} \lim\limits_{x\to 0}\left(\frac{3x+2}{x^2-2x+4}\right)& =\frac{3\left(0\right)+2}{\left(0\right)^2-2\left(0\right)+4}\\ \\ & =\frac{0+2}{0-0+4}\\ \\ & =\frac{2}{4}\\ \\ & =\frac{1}{2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \\ \end{align*}
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