# Definition of Limit of the Function

## Related Calculator: Limit Calculator

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.

To do this let's take points sufficiently close to 2 and evaluate corresponding y-values.

 x 1.9 1.99 1.999 2.0001 2.001 2.01 y=x^2 3.61 3.9601 3.996 4.0004 4.004 4.0401

We see that as x approaches 2, f(x) approaches 4, no matter we take points greater than 2 or less than 2. We write this fact as lim_(x->2)x^2=4.

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

In simple words this means that lim_(x->a)f(x)=L if we can make f(x) as close to L as we like by taking x sufficiently close to a.

We will also use following notation: f(x)->L as x->a.

From the definition it follows that we only care behaviour of function at points near a, not at a itself. This means that f(x) can even not be defined in a, but still have lim_(x->a)f(x).

For example, consider function f(x)=x^2 and g(x)={(x^2 if x!=2),(text(undefined) if x=2 ):}.

Note that lim_(x->2)f(x)=lim_(x->2)g(x)=4 even despite the fact that g(x) is not defined at 2. That's because we don't care about function at 2, we only interested in bevavior near 2.

Note that |x-a|<delta is equivalent to double inequality -delta<x-a<delta or a-delta<x<a+delta.

Let's do a couple of examples in trying to guess the limit. Note that this approach is not correct but it allows to better understand limits.

Example 2. Calculate lim_(x->1)(x+1).

Lets see what value does (x+1) approach as x approaches 1.

 x 0.9 0.99 0.999 1.0001 1.001 1.01 y=x+1 1.9 1.99 1.999 2.0001 2.001 2.01

So, as x approaches 1, f(x) approaches 2, thus lim_(x->1)(x+1)=2.

Example 3. Calculate lim_(x->1)(x-1)/(x^2-1).

Lets see what value does (x-1)/(x^2-1) approach when x approaches 1.

 x 0.9 0.99 0.999 1.0001 1.001 1.01 y=(x-1)/(x^2-1) (0.9-1)/(0.9^2-1)=0.5263 (0.99-1)/(0.99^2-1)=0.5025 0.5003 0.499975 0.49975 0.4975

So, as x approaches 1, f(x) approaches 0.5, thus lim_(x->1)(x-1)/(x^2-1)=1/2.

Example 4. Calculate lim_(x->0)cos(pi/x).

Lets see where cos(pi/x) goes when x approaches 0.

 x f(x) 1 cos(pi/1)=-1 0.1 sin(pi/0.1)=-1 0.001 sin(pi/0.001)=-1 0.0001 sin(pi/0.0001)=-1

It seems that lim_(x->0) cos(pi/x)=-1, but this is WRONG answer! The reason is that we took only values at which cos(pi/x)=-1.

Note that cos(pi/x)=-1 when x=1/n for any integer n.

But also cos(pi/x)=0 for infinitely many values of x that approach 0. In fact as x approaches 0, cosine take any value from interval [-1,1]. In other words cos(pi/x) oscillates infinitely many times as x approaches 0.

Since cos(pi/x) doesn't approach any fixed value then lim_(x->0)cos(pi/x) doesn't exist.

This example shows that we can guess wrong value if we take inappropriate values of x. It is also difficult to understand when to stop calculating values. However, we will give correct methods for calculating limits.

Example 5. Calculate lim_(x->0)1/x^2.

As x approaches 0, x^2 also approaches 0 so 1/x^2 becomes very large (for example when x=0.001 we have that 1/x^2=1000000). Thus, value of f(x) approach infinity without a bound, so lim_(x->0)1/x^2=oo.