Differential and Integral Calculus by Feliciano and Uy: Limit of a Function, Exercise 1.2, Problem 2

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

SOLUTION:

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

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

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

=\frac{4\cdot 3+2}{3+4}

=\frac{12+2}{7}

=\frac{14}{7}

=2

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