Differential and Integral Calculus| Feliciano and Uy| Limit of a Function| Exercise 1.2| Problem 2


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


Solution:

\displaystyle \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)}

\displaystyle =\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)}

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

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

\displaystyle =\frac{12+2}{7}

\displaystyle =\frac{14}{7}

\displaystyle =2


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