## Dodo1 2 years ago How do you find horizontal asymptote ?

1. ash2326

Suppose you have a function $y=f(x)$ Express x in terms of y, now find value of y for which x goes to infinity or - infinity That value of y is your horizontal asymptote

2. pottersheep

This website helped me a lot when I was learning about horizontal aymptotes. http://www.purplemath.com/modules/asymtote2.htm

3. pottersheep

Btw, your line CAN touch a horizontal aymptote. "Whereas vertical asymptotes are sacred ground, horizontal asymptotes are just useful suggestions. Whereas you can never touch a vertical asymptote, you can (and often do) touch and even cross horizontal asymptotes. Whereas vertical asymptotes indicate very specific behavior (on the graph), usually close to the origin, horizontal asymptotes indicate general behavior far off to the sides of the graph. "

4. ash2326

@Dodo1 do you understand?

5. Dodo1

This is a question A function is said to have a horizontal asymptote if either the limit at infinity exists or the limit at negative infinity exists. Show that each of the following functions has a horizontal asymptote by calculating the given limit. $\lim_{x \rightarrow \infty}10+\frac{ 3x }{ x^2-12x+3 }$

6. Dodo1

Yes, the basic concept! thank you i will take note.

7. pottersheep

Let me get my grade 12 notes

8. pottersheep

If y = [ax^n + ...............]/[Ax^N +.................] the horizontal asymptote = 0

9. pottersheep

Is the exponent on the bottom is greater than the one on the top, then it approaches zero

10. ash2326

Yes this is a way too, let's find the limit here $\lim_{x \rightarrow \infty}10+\frac{ 3x }{ x^2-12x+3 }$ Can you find the limit?

11. pottersheep

Because eventually, the number on the bottom will become HUGEEEEE compared to the top. And a number / a hugeeeeeeeee number is close to zero, a veryy small decimal!

12. pottersheep

That's how I understood it :)

13. Dodo1

ok, thsnk you potterssheep. :) how do i find limit?

14. ash2326

$\lim_{x \rightarrow \infty}10+\frac{ 3x }{ x^2-12x+3 }$ First divide the numerator and denominator by x $\lim_{x \rightarrow \infty}10+\frac{ 3\frac x x }{ \frac{x^2}{x}-12\frac x x +\frac 3 x}$ We'll get $\lim_{x\to \infty} 10+\frac 3 {x-12+\frac 3 x }$ 3/x =0 so we get $\lim_{x\to \infty} 10+\frac 3 {x-12}$ as $$x\to \infty$$ we get \$10+\frac 3 \infty$ $=10+0=>10$ so that's the limit

15. Dodo1

I see but why x-12= infinity?

16. ash2326

$x-12$ $\infty-12$ Infinity is a very big no. subtracting 12 won't change it $\infty-12 \longrightarrow \infty$

17. Dodo1

Oh i see thank you!! how about $\lim_{x \rightarrow -\infty} \frac{ 5-8x }{ 7+x }+\frac{ (6x^2+8) }{ (14x-12)^2 }$

18. Dodo1

Do I mutliply the ()2 first then add?

19. ash2326

$\lim_{x \rightarrow -\infty} \frac{ 5-8x }{ 7+x }+\frac{ (6x^2+8) }{ (14x-12)^2 }$ Let's split the limits $\lim_{x \rightarrow -\infty} \frac{ 5-8x }{ 7+x }$ First step divide by x , numerator and denominator $\lim_{x \rightarrow -\infty} \frac{\frac 5 x -8}{\frac 7 x +1}$ 1/x terms will become 0, when x goes to + or - infinity. We'll get $\frac{-8}{1}$ Do you get this?

20. Dodo1

OK, i got it so far.

21. Dodo1

:) its fun!

22. ash2326

Great, now the second part $\lim_{x \rightarrow -\infty}\frac{ (6x^2+8) }{ (14x-12)^2 }$ Let's expand the denominator $\lim_{x \rightarrow -\infty}\frac{ (6x^2+8) }{ (196x^2-336x+144) }$ Divide numerator and denominator by x^2, $\lim_{x \rightarrow -\infty}\frac{ (6+\frac8{x^2}) }{ (196-336\frac{x}{x^2}+\frac{144}{x^2}) }$Can you find the limit from here?

23. Dodo1

mmm, 6/(196-336)?

24. ash2326

x/x^2=1/x $x\to -\infty$$\frac 1 x \to 0, \frac 1 {x^2} \to 0$

25. ash2326

Try again now :)

26. Dodo1

6/196?

27. ash2326

yes, limit is the combined limit

28. Dodo1

whats combined limit?

29. ash2326

$-8+\frac 6 {196}$

30. Dodo1

Oh i see! - infinity and infinity are really matter? beacuse it seems that it does nt matter.?

31. ash2326

It does matter, in this question it doesn't

32. Dodo1

oh, not horizentally matter?

33. ash2326

nope

34. Dodo1

got it! thank you, i have other 4 questions but i will try and if i stuck can i ask you?

35. ash2326

Okay, try them. If I'm here, I'll help you But ask them in a new question. Close this