Lopital's rule

Definition 1

L'Hôpital's rule: under certain conditions, the limit of the ratio of functions whose variable tends to $ a $ is equal to the limit of the ratio of their derivatives, when $ x $ also tends to $ a $:

$ \ mathop (\ lim) \ limits_ (x \ to a) \ frac (f (x)) (g (x)) = \ mathop (\ lim) \ limits_ (x \ to a) \ frac (f "( x)) (g "(x)) $

Lopital's rule was discovered by the Swedish mathematician Johann Bernoulli, who then told Lopital about it in a letter. L'Hôtal published this rule in the first textbook on differential calculus in 1696 with his own authorship.

L'Hôpital's rule applies to expressions that reduce to uncertainties of the following form:

$ \ frac (0) (0) \ begin (array) (ccc) () & () & (\ frac (\ infty) (\ infty)) \ end (array) $

Instead of zero in the first expression, there can be some infinitesimal value.

In the general case, L'Hôpital's rule can be used if both the numerator and the denominator are at the same time zero or infinity.

Conditions under which the L'Hôpital rule can be applied:

• The condition is met under which the limits of the functions $ f (x) $ and $ g (x) $ as $ x $ tends to $ a $ are equal to each other and tend to zero or infinity: $ \ mathop (\ lim) \ limits_ (x \ to a) f (x) = \ mathop (\ lim) \ limits_ (x \ to a) g (x) = 0 $ or $ \ mathop (\ lim) \ limits_ (x \ to a) f (x) = \ mathop (\ lim) \ limits_ (x \ to a) g (x) = \ infty $;
• It is possible to get the derivatives $ f (x) $ and $ g (x) $ in the neighborhood of $ a $;
• The derivative of the function $ g (x) $ is not zero $ g "(x) \ ne 0 $ in a neighborhood of $ a $;
• The limit of the ratio of the derivatives of the functions $ f (x) $ and $ g (x) $, in the notation looks like $ \ mathop (\ lim) \ limits_ (x \ to a) \ frac (f "(x)) (g" ( x)) $ exists.

Proof of L'Hôpital's rule:

1. Let the functions $ f (x) $ and $ g (x) $ be given, and the equality of the limits is observed:
2. $ \ mathop (\ lim) \ limits_ (x \ to a + 0) f (x) = \ mathop (\ lim) \ limits_ (x \ to a + 0) g (x) = 0 $.
3. Let us extend the definition of the functions at the point $ a $. For this point, the following condition will be true:
4. $ \ frac (f (x)) (g (x)) = \ frac (f (x) -f (a)) (g (x) -g (a)) = \ frac (f "(c)) (g "(c)) $.
5. The value of $ c $ depends on $ x $, but if $ x \ to a + 0 $, then $ c \ to a $.
6. $ \ mathop (\ lim) \ limits_ (x \ to a + 0) \ frac (f (x)) (g (x)) = \ mathop (\ lim) \ limits_ (c \ to a + 0) \ frac (f "(c)) (g" (c)) = \ mathop (\ lim) \ limits_ (x \ to a + 0) \ frac (f "(c)) (g" (c)) $.

Algorithm for calculating the solution using L'Hôpital's rule

1. Checking the entire expression for uncertainty.
2. Check all the conditions stated above before further use of the L'Hôpital rule.
3. Checking the convergence of the derivative of the function to $ 0 $.
4. Rechecking for uncertainty.

Example # 1:

Find the limit:

$ \ mathop (\ lim) \ limits_ (x \ to 0) \ frac (x ^ (2) + 5x) (3x) $

Solution:

• The limit of the $ f (x) $ function is equal to the limit of $ g (x) $ and both of them are equal to zero: $ \ mathop (\ lim) \ limits_ (x \ to a) f (x) = \ mathop (\ lim) \ limits_ (x \ to 0) (x ^ (2) + 5x) = 0 $; $ \ mathop (\ lim) \ limits_ (x \ to a) g (x) = \ mathop (\ lim) \ limits_ (x \ to 0) (3x) = 0 $
• $ g "(x) = 3 \ ne 0 $ in the neighborhood of $ a $
• $ \ mathop (\ lim) \ limits_ (x \ to a) \ frac (f "(x)) (g" (x)) = \ mathop (\ lim) \ limits_ (x \ to 0) \ frac (2x +5) (3) $

$ \ mathop (\ lim) \ limits_ (x \ to 0) \ frac (x ^ (2) + 5x) (3x) = \ left \ langle \ frac (0) (0) \ right \ rangle = \ mathop ( \ lim) \ limits_ (x \ to 0) \ frac (\ left (x ^ (2) + 5x \ right) ") (\ left (3x \ right)") = \ mathop (\ lim) \ limits_ (x \ to 0) \ frac (2x + 5) (3) = \ frac (0 + 5) (3) = \ frac (5) (3) $

Example # 2:

Find the limit:

$ \ mathop (\ lim) \ limits_ (x \ to \ infty) \ frac (x ^ (3) -3x ^ (2) + 2x) (x ^ (3) -x) $

Solution:

Let us check the conditions for the applicability of the L'Hôpital rule:

• $ \ mathop (\ lim) \ limits_ (x \ to a) f (x) = \ mathop (\ lim) \ limits_ (x \ to \ infty) (x ^ (3) -3x ^ (2) + 2x) = \ infty $; $ \ mathop (\ lim) \ limits_ (x \ to a) g (x) = \ mathop (\ lim) \ limits_ (x \ to \ infty) (x ^ (3) -x) = \ infty $
• $ f (x) $ and $ g (x) $ are differentiable in a neighborhood of $ a $
• $ g "(x) = 6 \ ne 0 $ in the neighborhood of $ a $
• $ \ mathop (\ lim) \ limits_ (x \ to a) \ frac (f "(x)) (g" (x)) = \ mathop (\ lim) \ limits_ (x \ to \ infty) \ frac ( 3x ^ (2) -6x + 2) (3x ^ (2) -1) $

Let's write down the derivative and find the limit of the function:

$ \ mathop (\ lim) \ limits_ (x \ to \ infty) \ frac (x ^ (3) -3x ^ (2) + 2x) (x ^ (3) -x) = \ left \ langle \ frac ( \ infty) (\ infty) \ right \ rangle = \ mathop (\ lim) \ limits_ (x \ to \ infty) \ frac (\ left (x ^ (3) -3x ^ (2) + 2x \ right) " ) (\ left (x ^ (3) -x \ right) ") = \ mathop (\ lim) \ limits_ (x \ to \ infty) \ frac (3x ^ (2) -6x + 2) (3x ^ ( 2) -1) = \ left \ langle \ frac (\ infty) (\ infty) \ right \ rangle $

We repeat the calculation of the derivative until we get rid of the uncertainty:

$ \ mathop (\ lim) \ limits_ (x \ to \ infty) \ frac (\ left (3x ^ (2) -6x + 2 \ right) ") (\ left (3x ^ (2) -1 \ right) ") = \ mathop (\ lim) \ limits_ (x \ to \ infty) \ frac (6x-6) (6x) = \ left \ langle \ frac (\ infty) (\ infty) \ right \ rangle = \ mathop (\ lim) \ limits_ (x \ to \ infty) \ frac (\ left (6x-6 \ right) ") (\ left (6x \ right)") = \ frac (6) (6) = 1 $

Example No. 3:

Find the limit:

$ \ mathop (\ lim) \ limits_ (x \ to 0) \ frac (\ sin 5x) (x) $

Solution:

$ \ mathop (\ lim) \ limits_ (x \ to 0) \ frac (\ sin 5x) (x) = \ left \ langle \ frac (0) (0) \ right \ rangle = \ mathop (\ lim) \ limits_ (x \ to 0) \ frac (\ left (\ sin 5x \ right) ") (\ left (x \ right)") = \ mathop (\ lim) \ limits_ (x \ to 0) \ frac (5 \ cos 5x) (1) = 5 \ mathop (\ lim) \ limits_ (x \ to 0) \ cos 5x = 5 $

Example No. 4:

Find the limit:

$ \ mathop (\ lim) \ limits_ (x \ to \ infty) (1 + x ^ (2)) ^ (1 / x) $

Solution:

Let's logarithm the function:

$ \ ln y = \ frac (1) (x) \ ln (1 + x ^ (2)) = \ frac (\ ln (1 + x ^ (2))) (x) $

$ \ mathop (\ lim) \ limits_ (x \ to \ infty) \ frac (\ ln (1 + x ^ (2))) (x) = \ mathop (\ lim) \ limits_ (x \ to \ infty) \ frac (\ left [\ ln (1 + x ^ (2)) \ right] ") (x") = \ mathop (\ lim) \ limits_ (x \ to \ infty) \ frac (\ frac (2x) (1 + x ^ (2))) (1) = 0 $

Since the function $ ln (y) $ is continuous, we get:

$ \ mathop (\ lim) \ limits_ (x \ to \ infty) (\ ln y) = \ ln (\ mathop (\ lim) \ limits_ (x \ to \ infty) y) $

Hence,

$ \ ln (\ mathop (\ lim) \ limits_ (x \ to \ infty) y) = 0 $

$ \ mathop (\ lim) \ limits_ (x \ to \ infty) y = 1 $

$ \ mathop (\ lim) \ limits_ (x \ to \ infty) (1 + x ^ (2)) ^ (1 / x) = 1 $

We have already begun to deal with limits and their solution. Let's continue in hot pursuit and deal with the solution of the limits by L'Hôpital's rule... This simple rule can help you get out of the insidious and complex traps that teachers love to use in examples on tests in higher mathematics and calculus. The solution by the L'Hôpital rule is simple and quick. The main thing is to be able to differentiate.

L'Hôpital's rule: history and definition

In fact, this is not exactly L'Hôpital's rule, but the rule Lopital-Bernoulli... It was formulated by a Swiss mathematician Johann Bernoulli and the Frenchman Guillaume Lopital first published in his textbook infinitesimal in the glorious 1696 year. Can you imagine how people had to solve the limits with the disclosure of uncertainties before this happened? We are not.

Before proceeding with the analysis of L'Hôpital's rule, we recommend that you read an introductory article about limits in mathematics and methods for solving them. Often in the tasks one can find the formulation: find the limit without using L'Hôpital's rule. Also read about the techniques that will help you with this in our article.

If you are dealing with the limits of a fraction of two functions, be prepared: you will soon encounter an uncertainty of the form 0/0 or infinity / infinity. What does it mean? In the numerator and denominator, expressions tend to zero or infinity. At first glance, what to do with such a limit is completely incomprehensible. However, if you apply L'Hôpital's rule and think a little, everything falls into place.

But let us formulate the L'Hôpital-Bernoulli rule. To be perfectly precise, it is expressed by a theorem. L'Hôpital's rule, definition:

If two functions are differentiable in a neighborhood of the point x = a vanish at this point, and there is a limit on the ratio of the derivatives of these functions, then for NS aiming for a there is a limit of the ratio of the functions themselves, equal to the limit of the ratio of derivatives.

Let's write down the formula, and everything will immediately become easier. L'Hôpital's rule, formula:

Since we are interested in the practical side of the question, we will not present the proof of this theorem here. You will have to either take our word for it, or find it in any textbook on mathematical analysis and make sure that the theorem is correct.

By the way! For our readers, there is now a 10% discount on any kind of work

Disclosure of Uncertainties by L'Hôpital's Rule

What uncertainties can L'Hôpital's rule help with? Earlier we talked mainly about uncertainty. 0/0 ... However, this is far from the only uncertainty that can be encountered. Other types of uncertainties are:

Consider transformations that can be used to bring these uncertainties to the form 0/0 or infinity / infinity. Once converted, you can apply the L'Hôpital-Bernoulli rule and click the examples like nuts.

Uncertainty of the species infinity / infinity is reduced to the uncertainty of the form 0/0 simple conversion:

Let there be a product of two functions, one of which the first tends to zero, and the second - to infinity. We apply the transformation, and the product of zero and infinity turns into uncertainty. 0/0 :

To find limits with uncertainties like infinity minus infinity we use the following transformation leading to uncertainty 0/0 :

In order to use L'Hôpital's rule, you need to be able to take derivatives. We give below a table of derivatives of elementary functions, which you can use when solving examples, as well as the rules for calculating derivatives of complex functions:

Now let's move on to examples.

Example 1

Find the limit by L'Hôpital's rule:

Example 2

Calculate using L'Hôpital's rule:

An important point! If the limit of the second and subsequent derivatives of functions exists for NS aiming for a , then L'Hôpital's rule can be applied several times.

Find the limit ( n - natural number). For this we apply the L'Hôpital rule n once:

We wish you good luck in mastering mathematical analysis. And if you need to find the limit using L'Hôpital's rule, write an abstract according to L'Hôpital's rule, calculate the roots of a differential equation, or even calculate the tensor of inertia of a body, contact our authors. They will be happy to help you understand the intricacies of the solution.

There are various solution methods and formulas for solving the limits. But the fastest and easiest way, as well as universal, is the L'Hôpital method. In order to successfully use this wonderful simple way of calculating the limits, it is good enough to be able to find the derivatives of various functions. Let's start with theory.

Let us formulate L'Hôpital's rule. If:

• $ \ lim \ limits_ (x \ to a) f (x) = \ lim \ limits_ (x \ to a) g (x) = 0 \ text (or) \ infty $
• There are $ f "(a) \ text (and) g" (a) $
• $ g "(x) \ neq0 $
• There is $ \ lim \ limits_ (x \ to a) \ frac (f (x)) (g (x)) $

then there is $ \ lim \ limits_ (x \ to a) \ frac (f (x)) (g (x)) = \ lim \ limits_ (x \ to a) \ frac (f "(x)) (g" (x)) $

1. Substitute the point $ x $ in the limit
2. If $ \ frac (0) (0) \ text (or) \ frac (\ infty) (\ infty) $ is obtained, then we find the derivative of the numerator and denominator
3. Substitute the point $ x $ in the resulting limit and calculate it. If uncertainty is obtained, then we repeat paragraphs 2 and 3

Solution examples

To summarize: L'Hôpital's rule is a way and method thanks to which it is possible to disclose uncertainties of the form $ \ frac (0) (0) $ and $ \ frac (\ infty) (\ infty) $ when calculating limits. Its essence is that the limit of the ratio of functions is equal to the limit of the ratio of the derivatives of these functions.

• L'Hôpital's Rule and Uncertainty Disclosure
• Disclosure of uncertainties of the types "zero divided by zero" and "infinity divided by infinity"
• Disclosure of uncertainties of the form "zero times infinity"
• Disclosure of uncertainties of the types "zero to the power of zero", "infinity to the power of zero" and "one to the power of infinity"
• Disclosure of uncertainties of the form "infinity minus infinity"

L'Hôpital's Rule and Uncertainty Disclosure

Disclosure of uncertainties of the form 0/0 or ∞ / ∞ and some other uncertainties is greatly simplified using the L'Hôpital rule.

The essence L'Hôpital's rules consists in the fact that in the case when the calculation of the limit of the ratio of two functions gives uncertainties of the form 0/0 or ∞ / ∞, the limit of the ratio of two functions can be replaced by the limit of the ratio of their derivatives and, thus, a definite result can be obtained.

In general, L'Hôpital's rules mean several theorems that can be conveyed in the next one formulation.

Lopital's rule... If functions f(x) and g(x) are differentiable in some neighborhood of the point, with the possible exception of the point itself, and in this neighborhood

(1)

In other words, for uncertainties of the form 0/0 or ∞ / ∞, the limit of the ratio of two functions is equal to the limit of the ratio of their derivatives, if the latter exists (finite or infinite).

In equality (1), the value to which the variable tends can be either a finite number, or infinity, or minus infinity.

Uncertainties of other types can also be reduced to uncertainties of the 0/0 and ∞ / ∞ types.

Disclosure of uncertainties of the types "zero divided by zero" and "infinity divided by infinity"

Example 1. Calculate

x= 2 results in an uncertainty of the form 0/0. Therefore, we apply L'Hôpital's rule:

Example 2. Calculate

Solution. Substitution of a value in a given function x

Example 3. Calculate

Solution. Substitution of a value in a given function x= 0 results in an uncertainty of the form 0/0. Therefore, we apply L'Hôpital's rule:

Example 4. Calculate

Solution. Substitution of the x value equal to plus infinity into the given function leads to an uncertainty of the form ∞ / ∞. Therefore, we apply L'Hôpital's rule:

Comment. If the limit of the derivative ratio is an uncertainty of the form 0/0 or ∞ / ∞, then the L'Hôpital rule can be applied again, i.e. go to the limit of the ratio of the second derivatives, etc.

Example 5. Calculate

Solution. Find

Here L'Hôpital's rule is applied twice, since both the limit of the ratio of functions and the limit of the ratio of derivatives give an uncertainty of the form ∞ / ∞.

Example 6. Calculate

15. L'Hôpital's Rules *

Swiss mathematician Johann I Bernoulli(1667-1748) after successfully graduating from the University of Basel, traveling across Europe, in 1690 he came to Paris. In the literary salon of the philosopher Nicolas Malebranche (1638-1715), Johann meets the French mathematician Marquis Guillaume François Antoine de L'Hôpital (1661-1704). In the course of a lively conversation, Lopital was surprised at how easily, “as if playing,” the youth Bernoulli solved difficult problems according to the new calculus. Therefore, Lopital asked to give him several lectures. Lopital liked the oral conversations, and he began to receive materials in writing for a decent fee. Note that the now well-known “L'Hôpital's rule” for disclosing uncertainties was also passed on to him by Johannes. Already in 1696, L'Hôpital's famous treatise, “Introduction to the Analysis of the Infinitesimal for Understanding Curved Lines,” appeared. The second part of the course, outlined by Johann I Bernoulli, was published only in 1742 and was called “Mathematical lectures on the method of integrals and others; written for the famous Marquis of the Hospital; years 1691-1692 ”. In 1921, handwritten copies of the lectures, written by the hand of Johann I Bernoulli, were discovered, the originals of which were handed over to Lopital in 1691-1692. Of these, scientists unexpectedly discovered that Loptal in his "Analysis" almost did not deviate from the lectures of his young teacher.

Theorem (Cauchy). Let functions and be continuous on and differentiable on and. Then :

Proof. Consider the function

We choose so that all conditions of Rolle's theorem are satisfied, i.e. ...

According to Rolle's theorem, there is:

L'Hôpital's first rule

Definition. Let the functions be continuous on, differentiable in, and. Let be . Then the relation at is said to be an uncertainty of the form.

Theorem.

We apply Cauchy's theorem to the segment, where. Exists :

and therefore

This means that .

In the case when is infinite, inequality (1) is replaced by

depending on the sign. The rest of the proof does not change.

L'Hôpital's second rule

Definition. Let the functions be continuous and differentiable in, and. Let be . Then the relation at is said to be an uncertainty of the form.

Theorem. If under the specified conditions there is

Proof. Let of course. By choice: the interval satisfies the inequality

Let us define a function from the condition

at . We apply the Cauchy theorem to the segment. We get that there is:

For those for whom

Since it is arbitrarily small, then

In the case when inequality (2) is replaced by

and inequality (4) - on the inequality

which takes place for sufficiently close to a by virtue of (3).

The case is considered similarly.