How do you find the vertical, horizontal and slant asymptotes of: f(x)=sinx/(x(x^2-81))f(x)=sinxx(x2−81)?
1 Answer
It has a hole at
It has no slant asymptotes.
Explanation:
Given:
f(x) = (sin x)/(x(x^2-81)) = (sin x)/(x(x-9)(x+9))f(x)=sinxx(x2−81)=sinxx(x−9)(x+9)
Note that:
-
f(x)f(x) is undefined when the denominator is00 , i.e. whenx = 0x=0 ,x = -9x=−9 orx = 9x=9 . -
When
x=0x=0 then numerator is also00 , but we find:lim_(x->0) (sin x)/(x(x-9)(x+9)) = lim_(x->0) ((sin x)/x * 1/((x-9)(x+9))) = 1 * 1/(-81) = -1/81
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Sof(x) has a hole atx=0 . -
When
x=+-9 then numerator is non-zero, sof(x) has vertical asymptotes at these values. -
If
x is real-valued thenabs(sin x) <= 1 . Hence:lim_(x->oo) (sin x)/(x(x^2-81)) = 0
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andf(x) has a horizontal asymptotey=0
Footnote
Historically the line