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ash2326Best ResponseYou've already chosen the best response.1
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
 one year ago

pottersheepBest ResponseYou've already chosen the best response.0
This website helped me a lot when I was learning about horizontal aymptotes. http://www.purplemath.com/modules/asymtote2.htm
 one year ago

pottersheepBest ResponseYou've already chosen the best response.0
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. "
 one year ago

ash2326Best ResponseYou've already chosen the best response.1
@Dodo1 do you understand?
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
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. \[\lim_{x \rightarrow \infty}10+\frac{ 3x }{ x^212x+3 }\]
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
Yes, the basic concept! thank you i will take note.
 one year ago

pottersheepBest ResponseYou've already chosen the best response.0
Let me get my grade 12 notes
 one year ago

pottersheepBest ResponseYou've already chosen the best response.0
If y = [ax^n + ...............]/[Ax^N +.................] the horizontal asymptote = 0
 one year ago

pottersheepBest ResponseYou've already chosen the best response.0
Is the exponent on the bottom is greater than the one on the top, then it approaches zero
 one year ago

ash2326Best ResponseYou've already chosen the best response.1
Yes this is a way too, let's find the limit here \[\lim_{x \rightarrow \infty}10+\frac{ 3x }{ x^212x+3 }\] Can you find the limit?
 one year ago

pottersheepBest ResponseYou've already chosen the best response.0
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!
 one year ago

pottersheepBest ResponseYou've already chosen the best response.0
That's how I understood it :)
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
ok, thsnk you potterssheep. :) how do i find limit?
 one year ago

ash2326Best ResponseYou've already chosen the best response.1
\[\lim_{x \rightarrow \infty}10+\frac{ 3x }{ x^212x+3 }\] 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 \[\lim_{x\to \infty} 10+\frac 3 {x12+\frac 3 x }\] 3/x =0 so we get \[\lim_{x\to \infty} 10+\frac 3 {x12}\] as \(x\to \infty\) we get \\[10+\frac 3 \infty\] \[=10+0=>10\] so that's the limit
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
I see but why x12= infinity?
 one year ago

ash2326Best ResponseYou've already chosen the best response.1
\[x12\] \[\infty12\] Infinity is a very big no. subtracting 12 won't change it \[\infty12 \longrightarrow \infty\]
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
Oh i see thank you!! how about \[\lim_{x \rightarrow \infty} \frac{ 58x }{ 7+x }+\frac{ (6x^2+8) }{ (14x12)^2 }\]
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
Do I mutliply the ()2 first then add?
 one year ago

ash2326Best ResponseYou've already chosen the best response.1
\[\lim_{x \rightarrow \infty} \frac{ 58x }{ 7+x }+\frac{ (6x^2+8) }{ (14x12)^2 }\] Let's split the limits \[\lim_{x \rightarrow \infty} \frac{ 58x }{ 7+x }\] 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?
 one year ago

ash2326Best ResponseYou've already chosen the best response.1
Great, now the second part \[\lim_{x \rightarrow \infty}\frac{ (6x^2+8) }{ (14x12)^2 }\] Let's expand the denominator \[\lim_{x \rightarrow \infty}\frac{ (6x^2+8) }{ (196x^2336x+144) }\] Divide numerator and denominator by x^2, \[\lim_{x \rightarrow \infty}\frac{ (6+\frac8{x^2}) }{ (196336\frac{x}{x^2}+\frac{144}{x^2}) }\]Can you find the limit from here?
 one year ago

ash2326Best ResponseYou've already chosen the best response.1
x/x^2=1/x \[x\to \infty \]\[\frac 1 x \to 0, \frac 1 {x^2} \to 0\]
 one year ago

ash2326Best ResponseYou've already chosen the best response.1
yes, limit is the combined limit
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
Oh i see!  infinity and infinity are really matter? beacuse it seems that it does nt matter.?
 one year ago

ash2326Best ResponseYou've already chosen the best response.1
It does matter, in this question it doesn't
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
oh, not horizentally matter?
 one year ago

Dodo1Best ResponseYou've already chosen the best response.0
got it! thank you, i have other 4 questions but i will try and if i stuck can i ask you?
 one year ago

ash2326Best ResponseYou've already chosen the best response.1
Okay, try them. If I'm here, I'll help you But ask them in a new question. Close this
 one year ago
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