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ash2326
 2 years ago
Best ResponseYou've already chosen the best response.1Suppose 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

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

pottersheep
 2 years ago
Best ResponseYou've already chosen the best response.0Btw, 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. "

ash2326
 2 years ago
Best ResponseYou've already chosen the best response.1@Dodo1 do you understand?

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0This 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. \[\lim_{x \rightarrow \infty}10+\frac{ 3x }{ x^212x+3 }\]

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, the basic concept! thank you i will take note.

pottersheep
 2 years ago
Best ResponseYou've already chosen the best response.0Let me get my grade 12 notes

pottersheep
 2 years ago
Best ResponseYou've already chosen the best response.0If y = [ax^n + ...............]/[Ax^N +.................] the horizontal asymptote = 0

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

ash2326
 2 years ago
Best ResponseYou've already chosen the best response.1Yes 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?

pottersheep
 2 years ago
Best ResponseYou've already chosen the best response.0Because 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!

pottersheep
 2 years ago
Best ResponseYou've already chosen the best response.0That's how I understood it :)

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0ok, thsnk you potterssheep. :) how do i find limit?

ash2326
 2 years ago
Best 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

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0I see but why x12= infinity?

ash2326
 2 years ago
Best ResponseYou've already chosen the best response.1\[x12\] \[\infty12\] Infinity is a very big no. subtracting 12 won't change it \[\infty12 \longrightarrow \infty\]

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0Oh i see thank you!! how about \[\lim_{x \rightarrow \infty} \frac{ 58x }{ 7+x }+\frac{ (6x^2+8) }{ (14x12)^2 }\]

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0Do I mutliply the ()2 first then add?

ash2326
 2 years ago
Best 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?

ash2326
 2 years ago
Best ResponseYou've already chosen the best response.1Great, 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?

ash2326
 2 years ago
Best ResponseYou've already chosen the best response.1x/x^2=1/x \[x\to \infty \]\[\frac 1 x \to 0, \frac 1 {x^2} \to 0\]

ash2326
 2 years ago
Best ResponseYou've already chosen the best response.1yes, limit is the combined limit

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0Oh i see!  infinity and infinity are really matter? beacuse it seems that it does nt matter.?

ash2326
 2 years ago
Best ResponseYou've already chosen the best response.1It does matter, in this question it doesn't

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0oh, not horizentally matter?

Dodo1
 2 years ago
Best ResponseYou've already chosen the best response.0got it! thank you, i have other 4 questions but i will try and if i stuck can i ask you?

ash2326
 2 years ago
Best ResponseYou've already chosen the best response.1Okay, try them. If I'm here, I'll help you But ask them in a new question. Close this
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