Differential and Integral Calculus by Feliciano and Uy: Limit of a Function, Exercise 1.2, Problem 8

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

SOLUTION:

Plug in the value x=0.

\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}

 

Advertisements