Category: Limit of the Function

Definition of Limit of the Function

We already talked about limit of the sequence and since sequence is particular case of function then there will be similarity between sequence and function. And also we will extend some concepts.

Example 1. Let's investigate behavior of `y=x^2` near point `x=2`.

One-Sided Limits

Now we can extend concept of limit.

Definition. We write that `lim_(x->a^-)f(x)=L` and say "the limit of f(x), as x approaches a from the left, equals L" if for any `epsilon>0` there exists `delta>0` such that `|f(x)-L|<epsilon` when `|x-a|<delta` and `x>a`.

Sandwich Theorem

Fact. If `f(x)<=g(x)` when `x` is near `a` (except possibly at `a`) then `lim_(x->a)f(x)<=lim_(x->a)g(x)`.

This fact means that if values of `f(x)` are not larger than values of `g(x)` near `a`, then `f(x)` approaches not larger limit than `g(x)` as `x->a`.

Properties of the Limits

Now it is time to give properties of the limits that will allow to calculate limits easier.

Suppose that `c` is a constant and the limits `lim_(x->a)f(x)` and `lim_(x->a)g(x)` exist then following laws hold:

Limits Involving Infinity

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`.


Definition. The line `x=a` is called vertical asymptote of the curve `y=f(x)` if at least one of the following statements is true:

  1. `lim_(x->a)f(x)=oo`
  2. `lim_(x->a^+)f(x)=oo`
  3. `lim_(x->a^-)f(x)=oo`
  4. `lim_(x->a)f(x)=-oo`
  5. `lim_(x->a^+)f(x)=-oo`
  6. `lim_(x->a^-)f(x)=-oo`

For example y-axis (`x=0`) is a vertical asymptote of the curve `y=1/x^2` because `lim_(x->0)1/x^2=oo`.

Indeterminate Forms for Functions

In the note Limits Involving Infinity we saw that it is pretty easy to evaluate `lim_(x->0)(x+1)/x` because since `lim_(x->0)(x+1)=1` and `lim_(x->0)x=0` then division of 1 by very small number will give very large number, and so `lim_(x->0)(x+1)/x=oo`.