[SPLIT, MOVED] find hor. asymptote of f(x) = cos(x/x^3+1)

klpt

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i got stuck on a question of finding the horizontal asymptote of f(x) = cos(x/x3+1). I know in order to find horizontal asymptote (HA), we have to look function in the form of Numerator/Denominator, where Denominator suppose to be 0, and numerator should be some number.
 
I know in order to find horizontal asymptote (HA), we have to look function in the form of Numerator/Denominator....
This is true only for rational functions.

....where Denominator suppose to be 0, and numerator should be some number.
This is true only for vertical asymptotes.

i got stuck on a question of finding the horizontal asymptote of f(x) = cos(x/x3+1).
I will assume that the argument of the cosine is meant to be x/(x^3 + 1), rather than (1/x^2) + 1, which is the meaning of what you have posted.

To what value does x/(x^3 + 1) tend, as x gets arbitrarily large? (Use what they taught you for horizontal asymptotes for this.) What is the value of the cosine of that value? This will be your horizontal asymptote. ;)
 
actually question is simply "find the horizontal asymptote of f(x)=cos(x/x3+1)". i tried everything but cant get the answer , the answer is simply "y=1".
 
i tried everything but cant get the answer , the answer is simply "y=1".
Well, I'm guessing you didn't try "everything", since the steps outlined above would have gotten you the correct answer. Try those steps now, and see what you get. ;)
 
Thanks everyone for replying, i got the answer, i did some research and found that for finding trigonometric question H.A. (horizontal asymptote), we have to find the value of function by putting limit x->INFINITY and x-> -INFINITY. we dont have to follow those up written steps.
 
i did some research and found that for finding trigonometric question H.A. (horizontal asymptote), we have to find the value of function by putting limit x->INFINITY and x-> -INFINITY. we dont have to follow those up written steps.
Actually, that's exactly what "those up written steps" told you to do. ;)
 
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