help with lmitis question (1 Viewer)

[muttiah]

pg 370 4(a) - cambridge book year 11

Differenitage these functions using quotient rule. Use a table of values of y' to analyse any stationery pts. Find any asymptotes, then sketch each function:

y= x / x^2

y'=-(x^2 + 1) / (x^2 - 1)^2

so i done everything
and i got vertical asymptoes as 1,-1
horizontal asymptoes as 0.... and now i find that one of the x-intercepts is 0.. and in the book they have a line gonig through the origin.. how can they have this if the horizontal asymptote is 0. thank u bye!

 

[jake2.0]

The horizontal asymptote shows what the y-value would be as x approaches infinity, so it can be cut for lower values i.e. 0

 

[Riviet]

In other words, the curve CAN cut the asymptote at specific values, but as x approaches positive or negative infinity, they will approach the asymptote.

 

[muttiah]

hmm dats weird.. kk thanks for the help!

 

[followme]

lim..... [FONT=宋体] __x__[/FONT]
x->[FONT=宋体]±[/FONT]∞[FONT=宋体] .x[/FONT][FONT=宋体]²[/FONT]
[FONT=宋体]divide the top and bottom by x², and u'll get [/FONT]
lim ____1_
x->[FONT=宋体]±[/FONT]∞ __x
[FONT=宋体]
which is zero[/FONT]
[FONT=宋体].: the horizontal asymp is y=0
[/FONT]
[FONT=宋体]
[/FONT]