Limit without applying l'hopital's rule, $\lim_{x \rightarrow-\infty} \frac{|2x+5|}{2x+5}$. This is the question:
$$\lim_{x \rightarrow-\infty} \frac{|2x+5|}{2x+5}$$
I know the answer is $-1$, but can someone go through the steps and explaining it to me?
 A: You should better be aware of the definition of a modulus function. 
$$|x| = \left\{\begin{matrix}
x &  x > 0\\ 
0 & x = 0 \\ 
-x  & x<0 
\end{matrix}\right.$$
For this question, 
$$|2x+5| = \left\{\begin{matrix}
2x+5 &  2x+5 > 0\\ 
0 & 2x+5 = 0 \\ 
-(2x+5)  & 2x+5<0 
\end{matrix}\right.$$
Here, since, $x \to -\infty$ , that is $2x + 5 <0 $ , so, $|2x+5| = -(2x+5)$ and you get the limit as $-1$. 
To answer your question in specific that what will be its limit when $x$ tends to $-9/2$ or $-1/2$ . Well, here is the difference. 
$x \to -1/2 \implies 2x \to -1 \implies 2x + 5 \to 4 $ ($2x+4 >0$ )
So, $\lim_{x \to -1/2} \cfrac{|2x+5|}{2x + 5} = 1 $
While for $x\to -9/2 \implies 2x \to -9 \implies 2x + 5\to -4 $ ($2x+5 <0$)
So, $\lim_{x \to -9/2} \cfrac{|2x+5|}{2x +5} = - 1$ 
A: If you're confused by the function, hust set $2x+5 = t$, and you just have $\lim_{t \to - \infty} \frac{|t|}{t}$. Can you handle from here? 
A: The limits at infinity of a rational function is the limit of the ratio of the terms of highest degree.
Alternatively, with equivalents:
$2x+5\sim_{\pm\infty}2x$, $\,\lvert2x+5\rvert\sim_{-\infty}-2x$, hence
$$\frac{\lvert2x+5\rvert}{2x+5}\sim_{-\infty}\frac{-2x}{2x}=-1.$$
A: In $(-\infty,0)$ the function is identically the constant $-1$. The limit of a constant is this constant.
