How to find limits?

There is such a thing in mathematics as the limitfunction. To understand how to find the limits, remember the definition of the limit of a function: the function f (x) has the limit L at the point x = a if for each sequence of values of x that converges to 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, you can understand from the example. Suppose we have a function y = 1 / x. If we sequentially increase the value of x and look at what y is, we get all the decreasing values: for x = 10000 y = 1/10000; at x = 1,000,000 y = 1 / 1,000,000. Those. The more x, the less y. If x = ∞, y is so small that it can be considered equal to 0. Thus, the limit of the function y = 1 / x for x tending to ∞ is 0. This is written as follows:

lim1 / x = 0
x → ∞

The function limit has several properties that need to be remembered: this will make it much easier to solve the problems of finding the limits:

The limit of the sum 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 factor is taken as the limit sign: lim (Cx) = C lim x

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

lim (sin x) / x = 1 x → 0

In the article How to solve the limits, the methodology for solving such problems is described in detail. And we will consider several examples.

Solving examples for limits

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

Example 1

Lim (x-3) = lim (3-3) = 0
x → 3

Example 2

Lim [x² / (1-x)]. If we substitute x = ∞, we obtain
x → ∞
∞² / (1-∞) = ∞² / (-∞).

One infinity in the numerator and denominator is reduced:

∞ / (-1) = -∞. Hence,
Lim [x² / (1-x)] = -∞.
x → ∞

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

Example 3

Lim [(2х² - 3х - 5) / (1 + х + 3х²)]
x → ∞

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

Let us first try to divide both parts of the fraction into the higher degree - х²:

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

Hence, of all the terrible four-story fractions we still have:

Lim 2/3 = 2/3.

Answer:

Lim [(2х² - 3х - 5) / (1 + х + 3х²)] = 2/3
x → ∞

In this example, you could use the properties of the limits and convert the limit of the quotient to a private limit, and then represent the limits of the sum in the numerator and denominator as the sum of the 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 online service.