Asymptotes

Example 1. Find vertical asymptotes of `f(x)=x/(x-1)`.

In Limits Involving Infinity note we saw that if `lim_(x->a^+)f(x)=L!=0` and `lim_(x->a+)g(x)=0` then `lim_(x->a^+)(f(x)/g(x))=oo` (Law 8 for one-sided limits).

So, to find points where function approach infinity, we need to find points where denominator equals 0, and make sure that numerator doesn't equal 0 at this point.

The only point where denominator equals 0 is point 1, and numerator doesn't equal 0 at this point.

Let's find out one-sided limits at this point. When `x` approaches 1 from the right then `x-3` becomes very small positive value and `x` approaches 1. Thus, `lim_(x->1^+)(x)/(x-1)=oo` When `x` approaches 1 from the left then `x-1` becomes very small negative value and `x` approaches 1. Thus, `lim_(x->1^-)(x)/(x-1)=-oo`. So, we showed that both one-sided limits are infinite. In fact it was sufficiently to show that at least one of the limits is infinite.

Thus, `x=1` is vertical asympote.

Example 2. Find vertical asymptotes of `y=tan(x)`.

Recall that by definition `tan(x)=(sin(x))/(cos(x))`.

This means that there will be vertical asymptotes at points where `cos(x)=0`, i.e. points of the form `x=pi/2+pik,k in ZZ`.

Therefore, `tan(x)` has infinitely many asymptotes.

Example 3. Find vertical asymptotes of `f(x)= (x^2-x)/(x^2+x-2)`.

Since `x^2+x-2=(x+2)(x-1)` then denominator equals 0 when `x=-2` and `x=1`.

However, at point `x=1` numerator also equals 0. We need to additionally check this point.So, `lim_(x->1)(x^2-x)/(x^2+x-2)=lim_(x->1)(x(x-1))/((x+2)(x-1))=`

`=lim_(x->1)x/(x+2)=1/(1+2)=1/3`.

Thus, there is only one vertical asymptote `x=-2`. This example showed, that you should be very cautious: you should check whether numerator equals 0 at points where denominator equals 0.

Example 4. Find horizontal asymptotes of `y=1+1/x`.

Since `lim_(x->oo)(1+1/x)=1` then line `y=1` is horizontal asymptote.

In fact also `lim_(x->-oo)(1+1/x)=1`, but it suffices one limit to state that `y=1` is horizontal asymptote.

Example 5. Find horizontal asymptotes of `y=arctan(x)`.

Since `lim_(x->oo)arctan(x)=pi/2` and `lim_(x->-oo)arctan(x)=-pi/2` then there are two vertical asymptotes: `y=pi/2` and `y=-pi/2`.

Example 6. Find all asymptotes of `y=x+1/x`.

There are no horizontal asymptotes because `lim_(x->oo)(x+1/x)=oo` and `lim_(x->-oo)(x+1/x)=-oo`.

There is vertical asymptote `x=0` because `lim_(x->0^+)(x+1/x)=oo`.

This function also has oblique asymptote. Indeed, `m=lim_(x->oo)(x+1/x)/x=lim_(x->oo)(1+1/x^2)=1`.

So, `m=1`. Now, `b=lim_(x->oo)(x+1/x-1*x)=lim_(x->oo)1/x=0`. So, `b=0`.

Therefore, slant asymptote is `y=x`. On the figure to the right blue line is vertical asymptote `x=0`, green line is slant asymptote `y=x`.