Limits in Mathematics

Limits in mathematics

Limit of a function: Simple explanation

The expression to the right means that as x tends to some value a, f (x) approaches a limit L. This means that we can make the 'distance' of f (x) from the limit L (written | f (x) - L | ) as close as we like to 0 by choosing values for x closer and closer to a i.e. f (x) gets closer and closer to L.

It should be noted that f (a) may or may not equal L, depending on the continuity of the curve. If a curve is continuous at a then f( a ) = lim_x->af( x ) = L. (Basically if you could draw the graph in one line over some domain including both sides of the limit (in a finite amount of time without going directly up or down) it is continous)

The limit can be described by imagining the distance between our function f (x) and L as an 'error'. We will call this ε and use it to describe the maximum distance we want f (x) to be from our limit, L. f (x) could approach its limit from above or below and since we want to include both 'sides' we will take the absolute value so | f (x) - L | < ε .

The formal definition of a limit, is that for f (x) to approach a limit, L, as x --> a, we require that for all ε > 0 there exists a δ such that for all x satisfying 0 < | x - a | < δ, we have | f (x) - L | < ε.

This means all values of x satisfying 0 < | x - a | < δ will be closer to the limit L than the ε we choose. This is a a stronger condition than saying we can just find a single value for x that allows us to become as 'close' to our limit L as we like (the 'distance' | f (x) - L | as small as we like) because it means once the 'distance' between x and a is sufficiently small ( x becomes 'close' enough as necessary to a ) all of the the x values that are at least as 'close' to a will also be within the desired 'distance' | f (x) - L | ). This eliminates problems that may occur where for example equations could be approaching more than one limit, along with many others.

For simple problems, or those requiring no proof, it is often enough to simplify the problem then substitute in the value a for x to give L as on many occasions you will be dealing with curves that are continuous, or continuous around the point we are interested in.

For example, if we take L = lim_x->11 - x², this curve is continuous so

L = lim_x->1 1 - x² = 1 - (1)²

= 0

By the formal definition, we want | 1 - x² - 0 | < ε so if we solve this equation for x, we can determine δ:

|1 - x²| < ε

|-(x^2₎| < ε + 1

|x²| < ε + 1

|x| < √(ε + 1) = δ

This is our value for δ.

It is slightly more difficult where we have multiples of x in the denominator if we take L = lim_x->1(1 - x²) / x. This graph is discontinuous at x = 0 but not at x = 1, so we do not need to worry about that.

We recall we need f (x) to approach a limit, L, as x --> a, so we require that for all ε > 0 there exists a δ such that for all x satisfying 0 < | x - a | < δ, we have | f (x) - L | < ε.

In our case we want | (1 - x²) / x - 0 | < ε, now we will consider two cases:

x > 1

| (1 - x²) / x | < | 1 - x²|

lim_x->1 1 - x²= 0, so by the 'sandwich' theorem as | (1 - x²) / x | < | 1 - x²|for x > 1, we know
lim_x->1 (1 - x²) / x = 0 for x > 1.

Now we will consider 0 < x < 1 (We will not consider negative x, as we are not interested in this area, and will finding suitable upper bound curves easier).

We know that

| ( 1 - x²) / x | < | ( 1- x²) / x² | = | 1/x²- 1 | we use ( 1- x²) / x²as it has the same limit and is greater than ( 1 - x²) / x on 0 < x < 1. This allows us to simplify our problem so we may get ε that is not in terms of x.

lim_x->1| ( 1 - x²) / x | < lim_x->1| 1/x²- 1 | for 0 < x < 1 we just now need to show lim_x->11/x²= 1

We want | 1/x² - 1 | <ε:

| 1/x² | < ε + 1

| x² | < 1 / (ε + 1)

| x | < 1 / √ (ε + 1) = δ

We have again used the 'Sandwich' theorem to show that lim_x->1 (1 - x²) / x = 0 for 0 < x < 1, so we now see that this limit is approached on both sides, although we currently have two different δ. We can easily overcome this by setting δ = min { √(ε + 1) , 1 / √ (ε + 1) }, such that we always be 'close' enough in x for both x > 1 and 0 < x < 1.