# How to find the limits?

There is such a thing in mathematics as the limit function. To understand how to find the limits, you need to remember the definition of the limit of a function: the function f (x) has a limit L at the point x = a, if for each sequence of values of x converging to point a, the sequence of values of y approaches:

- L lim f (x) = L
- x → a

## The concept and properties of limits

What is the limit can be understood from the example. Suppose we have a function y = 1 / x. If we consistently increase the value of x and see what y is equal to, we will get all decreasing values: for x = 10,000, y = 1 / 10,000; at x = 1,000,000 y = 1 / 1,000,000. Those. the more x, the less y. If x = ∞, y will be so small that it can be considered equal to 0. Thus, the limit of the function y = 1 / x with x tending to ∞ is 0. This is written like this:

- lim1 / x = 0
- x → ∞

The limit of a function has several properties that must be remembered: this will greatly facilitate the solution of problems on finding limits:

- The sum limit is equal to the sum of the limits: lim (x + y) = lim x + lim y
- The limit of the product is the product of the limits: lim (xy) = lim x * lim y
- The limit of the quotient is equal to the quotient of the limits: lim (x / y) = lim x / lim y
- The constant multiplier is taken out of the limit sign: lim (Cx) = C lim x

For the function y = 1 / x, in which x → ∞, the limit is zero, as x → 0, the limit is ∞.

- lim (sin x) / x = 1 x → 0

The article How to solve limits describes in detail the methodology for solving such problems. And we will look at a few examples.

## Solving examples for limits

It is always necessary to begin to find the limits of functions by substituting into a function the value of x to which it tends.

### Example 1

- Lim (x-3) = lim (3-3) = 0
- x → 3

### Example 2

- Lim [х² / (1-х)]. If we substitute x = ∞, we get
- x → ∞
- ∞²/(1-∞) = ∞²/(-∞).

One infinity in the numerator and denominator is reduced:

- ∞ / (- 1) = -∞. So
- Lim [х² / (1-х)] = -∞.
- x → ∞

In these examples, everything is simple. However, the limits of functions are usually searched for at such values of x, which create an uncertainty of the type 0/0 or ∞ / ∞. Such uncertainties need to be disclosed.

### Example 3

- Lim [(2x² - 3x - 5) / (1 + x + 3x²)]
- x → ∞

We substitute x = ∞ and get infinity in the numerator and denominator, both there and there in the square. Hence, an uncertainty of type ∞ / ∞ is obtained.

Let us first try to divide both parts of the fraction by the senior degree - х²:

- Lim {[(2x² - 3x - 5) / x²] / [(1 + x + 3x²) / x²]} =
- x → ∞
- = Lim {[(2x² / x²) - (3x / x²) - (5 / x²)] / [(1 / x²) + (x / x²) + (3x² / x²)]} =
- x → ∞
- Lim {[2 - (3 / x) - (5 / x²)] / [(1 / x²) + (1 / x) + 3]}
- x → ∞
- With x = ∞ 3 / x = 0; 5 / х² = 0; 1 / x² = 0; 1 / x = 0.

So, of all the terrible four-story fraction we have left:

- Lim 2/3 = 2/3.

Answer:

- Lim [(2x² - 3x - 5) / (1 + x + 3x²)] = 2/3
- x → ∞

In this example, it was possible to use the properties of the limits and convert the limit of the quotient to the quotient of the limits, and then present the limits of the sum in the numerator and denominator as the sum of limits.

If you need to find the limit of a complex formula with which you do not know what to do, or just no time, you can use the service.