use f(x)= (3x+1)/ (Absolute value of x+2) to find lim x infinity, lim x - infinity, and all horizontal asymptotes??

Question

use f(x)= (3x+1)/ (Absolute value of x+2) to find lim x> infinity, lim x> - infinity, and all horizontal asymptotes??

anonymous · Answer

$$\lim_{x \rightarrow \infty}\frac{ 3x+1 }{ \left| x+2 \right|}=\lim_{x \rightarrow \infty}\frac{ 3x+1 }{ x+2 }=3$$$$\lim_{x \rightarrow -\infty}\frac{ 3x+1 }{ \left| x+2 \right|}=\lim_{x \rightarrow -\infty}\frac{ 3x+1 }{-(x+2) }=-3$$Those are precisely the horizontal asymptotes: y=3 and y=-3 
There is a vertical asymptote also in x=-2, what makes denominator=0

hero · Answer

$$f(x) = \frac{3x + 1}{|x+2|}$$

Separate into two 

$$f(x)=\frac{3x + 1}{x + 2}$$

$$f(x) = -\frac{3x + 1}{x + 2}$$


Re-write both as:

$$f(x) = 3 - \frac{5}{x + 2}$$
$$f(x) = \frac{5}{x + 2} - 3$$

To find the horizontal asymptotes, let y = f(x):
$$y = 3 - \frac{5}{x + 2}$$
$$y = \frac{5}{x + 2} - 3$$

Swap x and y:
$$x = 3 - \frac{5}{y + 2}$$
$$x = \frac{5}{y + 2} - 3$$

Then isolate y in each:
$$y = \frac{5}{3-x} - 2$$
$$y = \frac{5}{x + 3} -2$$

hero · Answer

In the first case we notice that $$x \ne 3$$ In the second case we notice that $$x \ne -3$$

The domain of the inverse function is the range of the original function. So 3 and -3 are the horizontal asymptotes of the original function.

anonymous · Answer

so whatever f(x) = will be the horizontal retricemptote??

anonymous · Answer

retricemptote*?

hero · Answer

@CarlosGP showed you how to find the horizontal asymptote by using limits. I showed you how to find it by using the inverse. The horizontal asymptotes occur where $$f(x) \ne 3$$$$f(x) \ne -3$$

hero · Answer

In the first case we notice that $$x \ne 3$$ In the second case we notice that $$x \ne -3$$

The domain of the inverse function is the range of the original function. So 3 and -3 are the horizontal asymptotes of the original function.