# Evaluate the limit? :  lim_(x rarr oo)(3x+1)/(|x|+2)

Nov 30, 2016

${\lim}_{x \rightarrow \infty} \frac{3 x + 1}{| x | + 2} = 3$

#### Explanation:

As $x \rightarrow \infty \implies x > 0$ so:
$\setminus \setminus \setminus \setminus \setminus {\lim}_{x \rightarrow \infty} \frac{3 x + 1}{| x | + 2} = {\lim}_{x \rightarrow \infty} \frac{3 x + 1}{x + 2}$
$\therefore {\lim}_{x \rightarrow \infty} \frac{3 x + 1}{| x | + 2} = {\lim}_{x \rightarrow \infty} \frac{3 x + 1}{x + 2} \cdot \frac{\frac{1}{x}}{\frac{1}{x}}$
$\therefore {\lim}_{x \rightarrow \infty} \frac{3 x + 1}{| x | + 2} = {\lim}_{x \rightarrow \infty} \frac{\frac{3 x + 1}{x}}{\frac{x + 2}{x}}$
$\therefore {\lim}_{x \rightarrow \infty} \frac{3 x + 1}{| x | + 2} = {\lim}_{x \rightarrow \infty} \frac{3 + \frac{1}{x}}{1 + \frac{2}{x}}$

As $x \rightarrow 8 \implies \frac{1}{x} \rightarrow 0$, Hence
$\setminus \setminus \setminus \setminus \setminus {\lim}_{x \rightarrow \infty} \frac{3 x + 1}{| x | + 2} = \frac{3 + 0}{1 + 0} = 3$

We can verify this by looking at the graph of $y = \frac{3 x + 1}{| x | + 2}$

graph{(3x+1)/(|x|+2) [-5, 20, -3.5, 3.5]}