Question

if f(x)=(sinx)/(x), how can i find the vertical asymptote?

Answer 1

Where is the denominator zero? Or is it? Are you SURE it has one?

Answer 2

That has no vertical asymptote.

Answer 3

You find vertical asymptotes by solving for the denominator, but this one is just x.

Answer 4

@ridoleor By your description, why isn't x = 0 a vertical asymptote?

Answer 5

because i thought that for finding vertical asymp., you had to set the denominator equal to zero, thus "x=0" but idk...it's for my calculus BC homework...

Answer 6

If x = 0 in the denominator and the num. is a basic trig function (sin x, cos x, tan x) it wil only have an asymptote if the equation was cos x /x

Answer 7

I don't know why it works like that, but it does. sin x/ x has no vert. asymptote

Answer 8

is there a way to justify it?

Answer 9

The only way really to justify why the rule doesn't work is because that equation doesn't have an infinite limit (it's 1),

Answer 10

oh, okay, thx(:

Answer 11

\[\lim_{x \rightarrow 0} =\infty \] for it to work

Answer 12

Quite unsatisfying.

The "rule" is NOT, "Denominator = 0, then you have a vertical asymptote."
The "rule" IS, "Denominator = 0, then you MIGHT have a vertical asymptote. Check the numerator."

Why not prove it, rather that not know?

1) x is not in the Domain. There will be no substitution.

2) f(0.1), f(0.01), f(0.001), f(0.0001), f(0.00001), ... It certainly doesn't LOOK like an asymptote.

3) \(\lim\limits_{x\rightarrow 0}\dfrac{sin(x)}{x} = 1\), which is a well-known result or it is easy enough to prove (There are other ways, but l'Hospital's rule makes it trivial)

4) x = 0 is the ONLY candidate for a vertical asymptote. Since it isn't actually a vertical asymptote, then there isn't one.

Answer 13

ohhhh, okay i think i'm getting it know, thx (: