Question

I need to look for the horizontal asymptote of
(sin x)/(1 + cos x)...How I do this? :/

Answer 1

cosx=-1
x=pie

Answer 2

there are undefined when 1+cosx=0
this is horyzontal asymptote

Answer 3

Do you know how can I resolve this using the limit x->infinity equation?

Answer 4

this is what I'm talking about http://www.wolframalpha.com/input/?i=lim+x-%3Einf+%28sen%28x%29%29%2F%281%2B+cos%28x%29%29

Answer 5

I need to find the asymptote using definition of limit when x-> infinity

Answer 6

\[\lim_{x \rightarrow \infty}\frac{sinx}{1+cosx}*\frac{1-cosx}{1-cosx}=\lim_{x \rightarrow \infty}\frac{sinx*(1-cosx)}{1-\cos^2x}\]
\[=\lim_{x \rightarrow \infty}\frac{sinx(1-cosx)}{\sin^2x}=\lim_{x \rightarrow \infty}\frac{1-cosx}{sinx}=\lim_{x \rightarrow \infty}(cscx-cotx)\]
there is no horizontal asymptote

Answer 7

you can find the vertical asymptotes by solving this for x: 1+cosx=0

Answer 8

okay thanks!

Answer 9

do you know any other method of how I can find horizontal asymptotes of
(x+1)^3 / (x-1)^2 ??
without using L'Hospital

Answer 10

using lim x->infinity....but not using L'Hospital

Answer 11

ok do you remember
\[\lim_{x \rightarrow \infty}\frac{1}{x^r}=0, r \in \mathbb{Z}\]

Answer 12

yes

Answer 13

\[\lim_{x \rightarrow \infty }\frac{(x+1)^3}{(x-1)^2}=\lim_{x \rightarrow \infty}\frac{x^3+3x^2+3x+1}{x^2-2x+1}\]
now divide top and bottom by x^2
\[\lim_{x \rightarrow \infty}\frac{x+3+\frac{3}{x}+\frac{1}{x^2}}{1-\frac{2}{x}+\frac{1}{x^2}}\]

Answer 14

top goes to infinity
bottom goes to 1
so this means there is no horizontal asymptote

Answer 15

now we didnt have to multiply that junk out i just thought it would be easier to see

Answer 16

if you have polynomial/polynomial
degree of top is equal to degree of bottom the horizontal asymptote is
y=coefficient of the leading term (highest exponent) on top over
coefficient of the leading term (highest exponent) on bottom
if the degree on top is higher than the degree on bottom like this one we had up here
then there is no horizontal asymptote
if the degree on top is lower than the degree on bottom then the horizontal asymptote
is y=0

Answer 17

PERFECT! :D...thanks!!!

Answer 18

so we have (3x^2+1)/(6x^2-9)
then horizontal asymptote is y=1/2

Answer 19

if we have
that is an example i made up k?

Answer 20

so if we have (x-1)^2/(x+1)^3
then the horizontal asymptote is y=0

Answer 21

what do you think the horiztontal asymptote is of
(2x-1)^3/(x-1)^2

Answer 22

degree on top>degree on bottom
so there is not one

Answer 23

i see

Answer 24

(2x-1)^3/(x-1)^3
this one?

Answer 25

first one...both have same exponents, so the asymptote is their coefficients

Answer 26

my first example to you yes
i said y=1/2
or you could have said y=3/6

Answer 27

okay

Answer 28

top degree > bottom degree = no asymptote
top degree < bottom degree = asymptote = 0
equal degree on denominator and numerator = leading coefficients

Answer 29

so my last example the degrees are the same right?
so on top you have after expanding it you have 8x^3 + gibberish
on bottom you have after expanding it you have x^3 + gibberish
so the horizontal asymptote is y=8

Answer 30

thanks! :D

Answer 31

right

Answer 32

leading coefficient of top
---------------------
leading coefficicent of bottom


if the degree of top=degree of bottom

Answer 33

yes

Answer 34

:)

Answer 35

very very helpful!