Dividing Functions Questions

CanadianEh

Hi there,
Quick question. For F(X)= X/Sin(X), is there a hole at X=0?

Thanks.

Bohrok

What do you get when plugging 0 into F(X) ?

tiny-tim

Hi CanadianEh!

At x = 0, obviously, it's 0/0, which is undefined (it's known as an "indeterminate form"), so yes in that sense there's a hole …

of course, F(x) does tend to a limit at as x -> 0

CanadianEh

0/Sin 0 = undefined.

So basically, there's my answer. There is a hole at x=0. There is also an oblique asymptote of f(x)=x, correct?

CanadianEh

Thanks so much! Can you help me explain why there is an oblique asymptote?

tiny-tim

uhh?

wot's an oblique asymptote?

CanadianEh

When a linear asymptote is not parallel to the x- or y-axis, it is called either an oblique asymptote or equivalently a slant asymptote.

In the graph of X/Sin(X), there appears to be an asymptote at y=x

chroot

The function continues to have a defined value as you get arbitrarily close to zero, thus the limit as x->0 is defined. The function itself is undefined only exactly at zero.

- Warren

Bohrok

Try graphing x/sin(x) and you'll only see vertical asymptotes when the denominator, or sin(x), is 0.
As far as I know, a rational function P(x)/Q(x) where P and Q are polynomials has an oblique asymptote only when the degree of the numerator is one larger than that of the denominator. In x/sin(x) you have a transcendental function in the denominator.

CanadianEh

Ok, so NO oblique asymptote, correct?

tiny-tim

Still totally confused as to why this is called an asymptote instead of a tangent.

Anyway I can't see how it's slanting ……

what is lim_{x -> 0} x/sinx ?

Bohrok

That's right.


A slant asymptote

tiny-tim

So that's only at infinity?

Bohrok

and also negative infinity if the domain goes there too.

HallsofIvy

tiny-tim, the word "asymptote" was wrong here. He intended "tangent" as you suggested. Because there is a "hole" at x= 0, there is no tangent there.