Limit without applying l'hopital's rule, $\lim_{x \rightarrow-\infty} \frac{|2x+5|}{2x+5}$.

Question

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?

Answer 1

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, $\lim_{x \to -9/2} \cfrac{|2x+5|}{2x +5} = - 1$

Answer 2

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?

Answer 3

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 $$\frac{\lvert2x+5\rvert}{2x+5}\sim_{-\infty}\frac{-2x}{2x}=-1.$$

Answer 4

In $(-\infty,0)$ the function is identically the constant $-1$. The limit of a constant is this constant.