Limits Involving Infinity

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Now it is time to talk about limits that involve special symbol `oo`.

First we talk about infinite limits.

Definition. We write that `lim_(x->a)f(x)=oo` (`lim_(x->a)f(x)=-oo`) if for any `E>0` there exists `delta>0` such that `f(x)>E\ (f(x)<-E)` when `|x-a|<delta`.

This definition says that `lim_(x->a)f(x)=oo` means the following: when `x` approaches `a` `f(x)` increases (decreases) without a bound and can take very large number.

Once again symbol `oo` is just used to denote very large number. We can't perform arithmetic operations on such numbers.

Note, that we asume that `lim_(x->a)f(x)=oo` exists, i.e. `f(x)` approach very large number (infinity), denoted by `oo`. Unlike `lim_(x->0)sin(1/x)` doesn't exist, because function doesn't approach any value.

So, we use symbol `oo` to denote the fact that function approaches very large number, not the fact that limit doesn't exist, i.e. function doesn't approach any number.

Example 1. Find `lim_(x->0) 1/x^2`.infinite limits

When `x` approaches 0, `x^2` also approaches 0, so, `1/x^2` approaches very large value (for example, `1/(0.001)^2=1000000`), thus `lim_(x->0)=1/x^2`.

Also it can be easily seen that `lim_(x->0)-1/x^2=-oo`.

Similarly, we can define infinite one-sided limits: `lim_(x->a^-)f(x)=oo`, `lim_(x->a^+)f(x)=oo`, `lim_(x->a^-)f(x)=-oo`, `lim_(x->a^+)f(x)=-oo`.

Example 2. Find `lim_(x->1)1/(x-1)`.infinite one sided limits

When we approach 1 from the left, then `x-1` is small positive number and `1/(x-1)` is large negative number. Therefore, `lim_(x->1^-)1/(x-1)=-oo`.

When we approach 1 from the right, then `x-1` is small positive number and `1/(x-1)` is large positive number. Therefore, `lim_(x->1^+)1/(x-1)=oo`.

Since `lim_(x->1^-)1/(x-1)!=lim_(x->1^+)1/(x-1)` then `lim_(x->1)1/(x-1)` doesn't exist.

Limits at Infinity

Above we let `x` approach some number and the result was that values of `f(x)` became arbitrary large (positive or negative).

Now we let `x` become arbitrary large (positive or negative) and see what happens to `f(x)`.

Example 3. Find `lim_(x->oo)(3x^2-1)/(x^2+1)` by guessing.limits at infinity

If `x=100` then `f(100)=(3*100^2-1)/(100^2+1)~~2.9996` and if `x=1000` then`f(1000)=(3*1000^2-1)/(1000^2+1)~~2.999996`.

As `x` grows larger and larger values of `f(x)` become more and more close to 3.

We can make `f(x)` as close to 3 as we like by choosing suffciently large `x`.

Thus, `lim_(x->oo)(3x^2-1)/(x^2+1)=3`.

Similarly we can show that `lim_(x->-oo)(3x^2-1)/(x^2+1)=3`.

Definition. We say that `lim_(x->oo)f(x)=L\ (lim_(x->-oo)f(x)=L)` if for every number `epsilon>0` there exists number `M>0` such that `|f(x)-L|<epsilon` when `x>M\ (x<-M)`.

This means that we can make `f(x)` as close as we like to `L` by taking sufficiently large `x` (positive or negative).

Example 4. Find `lim_(x->oo)1/(x-1)` and `lim_(x->-oo)1/(x-1)`.

Observe that when `x` is large `1/(x-1)` is small. Therefore, we can make `1/(x-1)` as close to 0 as we like by taking `x` sufficiently large (positive or negative). Therefore, `lim_(x->oo)1/(x-1)=lim_(x->-oo)1/(x-1)=0`. You can see graph of function in example 2.

Example 5. Evaluate `lim_(x->0^-)e^(1/x)`.

Let `t=1/x` then `t->-oo` as `x->0^-` . Therefore, `lim_(x->0^-)e^(1/x)=lim_(t->-oo)e^t=0`.

Again, note that `oo` is not a number, it is just the way to show that some value increases (or decreases) without a bound.

Example 6. Find `lim_(x->oo)cos(x)`.

As `x` increases value of `cos(x)` oscillates infinitely many times between -1 and 1. Thus, `lim_(x->oo)cos(x)` doesn't exist.

Infinite Limits at Infinity

Notation `lim_(x->oo)f(x)=oo` means that `f(x)` becomes large as `x` becomes large. Similar meaning have following notations: `lim_(x->-oo)f(x)=oo` , `lim_(x->oo)f(x)=-oo` , `lim_(x->-oo)f(x)=-oo` .

For example, `lim_(x->oo)e^x=oo` , `lim_(x->-oo)x^2=oo` , `lim_(x->oo)-e^x=-oo` and `lim_(x->-oo)x^3=-oo`.

Most of the properties of the limits hold for infinite limits. Now, let's formulate a couple of laws involving infinity.

Properties of infinite limits.

Suppose `a` is number (can be infinity) and `L` is finite number.

Law 1. If `lim_(x->a)f(x)=L` and `lim_(x->a)g(x)=oo` then `lim_(x->a)(f(x)+g(x))=oo`.

Law 2. If `lim_(x->a)f(x)=L` and `lim_(x->a)g(x)=oo` then `lim_(x->a)(f(x)-g(x))=-oo`.

Law 3. If `lim_(x->a)f(x)=L!=0` and `lim_(x->a)g(x)=oo` then `lim_(x->a)(f(x)g(x))=oo` if `L>0` and `-oo` if `L<0.`

Law 4. If `lim_(x->a)f(x)=L` and `lim_(x->a)g(x)=+-oo` then `lim_(x->a)(f(x)/g(x))=0`.

Law 5. If `lim_(x->a)f(x)=L` and `lim_(x->a)g(x)=oo` then `lim_(x->a)((g(x))/(f(x)))=oo` if `L>0` and `-oo` if `L<0`.

Law 6. If `lim_(x->a)f(x)=oo` and `lim_(x->a)g(x)=oo` then `lim_(x->a)(f(x)+g(x))=oo`.

Law 7. If `lim_(x->a)f(x)=oo` and `lim_(x->a)g(x)=-oo` then `lim_(x->a)(f(x)-g(x))=oo`.

Law 8. If `lim_(x->a)f(x)=L!=0` and `lim_(x->a)g(x)=0` then `lim_(x->a)(f(x)/g(x))=oo` if `L>0` and `-oo` if `L<0`.

Law 9. If `lim_(x->a)f(x)=L!=0` and `lim_(x->a)g(x)=0` then `lim_(x->a)(g(x)/f(x))=0`.

Law 10. If `lim_(x->a)f(x)=oo` then `lim_(x->a)cf(x)=oo` where `c!=0` is some constant.

Note, that these properties also hold for one-sided limits.

It is easy to believe that these properties are true.

Indeed, though we can't perform arithmetic operations over infinite values, but it is clear that sum of large numbers is large number again (Law 1), product of large numbers is again large number (Law 3), if we divide large number by very small number we will obtain again large number (Law 4) etc.

Example 7. Find `lim_(x->2)(1-x)/(x-2)`.

Since `lim_(x->2)(1-x)=-1` and `lim_(x->2)(x-2)=0` then by Law 8 `lim_(x->2)(1-x)/(x-2)=-oo`.

Example 8. Find `lim_(x->oo)(1-1/x)/(x)`.

Since `lim_(x->oo)(1-1/x)=1` and `lim_(x->oo)x=oo` then by Law 4 `lim_(x->oo)(1-1/x)/x=0`.