How do you find the vertical, horizontal and slant asymptotes of: #f(x)=sinx/(x(x^281))#?
1 Answer
Answer:
It has a hole at
It has no slant asymptotes.
Explanation:
Given:
#f(x) = (sin x)/(x(x^281)) = (sin x)/(x(x9)(x+9))#
Note that:

#f(x)# is undefined when the denominator is#0# , i.e. when#x = 0# ,#x = 9# or#x = 9# . 
When
#x=0# then numerator is also#0# , but we find:#lim_(x>0) (sin x)/(x(x9)(x+9)) = lim_(x>0) ((sin x)/x * 1/((x9)(x+9))) = 1 * 1/(81) = 1/81#
#""#
So#f(x)# has a hole at#x=0# . 
When
#x=+9# then numerator is nonzero, so#f(x)# has vertical asymptotes at these values. 
If
#x# is realvalued then#abs(sin x) <= 1# . Hence:#lim_(x>oo) (sin x)/(x(x^281)) = 0#
#""#
and#f(x)# has a horizontal asymptote#y=0#
Footnote
Historically the line