Question

Why is the vertical asymptote for `f(x)=sqrt((x-3)/x)` x=0 when the domain is (-infinity,0)`uu`(3,+infintiy)?

I need to find the asymptote for this function but I don't understand why the answer is x=0.

Our teacher told us to calculate lateral limits for, in this case, 0 and 3 and find if either is infinite as x approaches from right and left. But if that's the case then what happens when calculating the limits for x approaches 3 is it undefined?

Answer 1

See below.

Explanation:

Vertical asymptotes occur where the function is undefined. In this case at `bb(x=0)`. This give division by zero.

If we take the left and right limits as x approaches zero:

`lim_(x->0^+)sqrt((x-3)/x)` undefined for real numbers.

`lim_(x->0^-)sqrt((x-3)/x)=oo` ?

? This confirms an asymptote at `x=0`.

as `x` approaches `3`:

`lim_(x->3^+)sqrt((x-3)/x)=0`

`lim_(x->3^-)sqrt((x-3)/x)` undefined for real numbers.

If we plug `3` into the function:

`sqrt(((3)-3)/(3))=0`

So the function is defined at `x=3`

This is one of those cases where if we solve the limit by plugging in `3`, we conclude that:

`lim_(x->3)sqrt((x-3)/x)=0`

But we can see from the graph, that the left limit does not exist, as well as trying a value to the left of `3`:

`sqrt(((2.99)-3)/(2.99))=sqrt(-0.3344481605e-2`

This is the square root of a negative number.

So the domain should be:

`(-oo,0)uu[3,oo)`

Note the inclusion of `bb3`. The function is defined here.

graph{y=sqrt((x-3)/x) [-12.66, 12.65, -6.33, 6.33]}

In the question you are asking why the asymptote is `x=0`.

Remember that an asymptote is a line that a curve approaches, it never reaches it. The `x` value at the asymptote is never in the domain of the function, because the function is always undefined there.