Practical lesson 2. L’Hospital’s rule
The lesson is devoted to L’Hospital’s rule. What are we going to talk about? Of course, about calculating limits. And what a plus! We may even forget something: formulas of special limits, for example, and L’Hospital 's rule will help us.
Let's look at the first example (watch the video).
First, where do we start calculating any limit? This is not even related to the L’Hospital’s rule, just calculating limits. We start substituting instead of x the value to which it tends - zero. arcsin 0 is 0, cos 0 is 1, and we get an uncertainty of 0/0. We can't tell the answer right away. You can try to solve this by applying the first special limit and its consequences, but the topic of the lesson is the L’Hospital’s rule. That’s why we write down that the limit, x tends to 0.
Next, we find the derivative of the numerator, be careful. According to the rules for differentiating a complex function (watch the video). So, the derivative of the numerator was found, we go to the denominator. The denominator will contain the derivative of the denominator. Do not confuse, sometimes an error occurs, we find the derivative of the fraction. No! In the numerator we write the derivative of the numerator, in the denominator - the derivative of the denominator (watch the video). So, let's see what the answer is now.
So, x tends to zero, you don't need to convert – we count: 1/1, and this fraction became 1, multiply by 2, and here we get 2. And in the denominator we get 0, but the most important thing is that there is no uncertainty, and we write the answer: a number that is not equal to zero and divided by zero, the theory of limits gives us the answer ∞. So, the problem is solved. Look, we didn't have to know the formulas, the consequences of the first special limit, but only the knowledge of the L’Hospital’s rule.
Let's move on to the next example (watch the video). Here, I must say, it is not even clear what transformations to do if we did not have L’Hospital’s rule. Let's substitute 0 instead of x. We start calculating: sin 0 is 0, so the numerator will be 0 and the denominator will also be 0. We got the uncertainty 0/0. We begin to apply L’Hospital’s rule: lim at x tends to 0, however, we remove the plus sign - it is written with a plus sign here, but this is absolutely not important, we write that it simply tends to zero.
We will calculate: cosx-2cos 2x (the derivative of the sine is also the cosine of the same argument, multiplied by the derivative of the internal function (by 2)). In the denominator, we get 1-cosx. Every time we look at what we got. Substitute x 0 instead: cos 0 is one, 2-2=0, and in the denominator 1-1=0. Look, it looks like nothing happened. The uncertainty has not disappeared, it should not frighten us and should not stop us.
Let's try again to apply L’Hospital's rule. So, we write: lim at x tends to 0, in the numerator we find the derivative of the numerator: (the derivative of cos is sin) - 2sinx (here there will be a plus, do not forget, here the derivative of the internal function will give us another 2) +4sin2x, in the denominator. The derivative of cos is sin, so we write sinx. So, we substitute 0 instead of x: sin 0 is 0, we get the uncertainty 0/0. Well, I think that's where we'll end up finally applying L’Hospital’s rule.
Let's check: the derivative of sin is cos, write in the numerator: - 2cosx+8cos2x, and the denominator will be cosx, which at point 0 will no longer be 0. There is no absolute uncertainty, it remains to count: cos0 is 1, -2+8 is 6, 6:1 is 6. the Answer is received.
Let's watch what else might be interesting (watch the video). But here is just a one-way limit, and the situation is completely new for us. That’s all of a sudden? Why does x tend to 0 on the right? Because the logarithm exists only when x>0, then x cannot be less than zero.
Another feature: the limit of x is 0, and the limit of ln x is -∞. How do we find that out (watch the video)? Schematically we draw a graph of the logarithm and see that at point 0 it is equal to minus infinity. We are dealing with uncertainty, 0×∞. L’Hospital’s rule does not deal with such uncertainty. What will we do? We will represent this product as a fraction.
So, we will leave ln x in the numerator, and move x to the denominator, and this will become a fraction 1/x. So, x tends to the point 0 on the right. Let's check what happened to the uncertainty: ln tends to ∞, and the fraction 1/x (1/0) also tends to ∞. This is something that we need to apply L’Hospital 's rule. We write: lim at x tends to 0 on the right, in the numerator we write the derivative of the numerator: 1/x, in the denominator - the derivative of the denominator: -1/x^2.
Here, the easiest way is to simply convert this expression like this: lim at x tends to 0, if we divide the fraction by a fraction, we get: (minus can be taken out of the limit) x^2/x. It is clear that after reducing by x, we get 0 in the limit. The problem is solved. So, L’Hospital’s rule allowed us to even somehow deal with the uncertainty of 0×∞.
Let's look at other situations (watch the video). So, the limit of the difference. It would not seem to be about L’Hospital’s rule at all. If we substitute 1 instead of x, we calculate: 1/0 is ∞, ln 1 is 0, 1/0 is ∞again. This is uncertainty ∞-∞. What are we going to do here? Let's take and bring fractions to a common denominator. So, we got the limit, the denominator is (x-1)×ln x, the numerator is xln x –x +1, when x tends to 1. So, let's try to calculate, what if everything is fine. So, ln 1 is 0, -1+1=0, in the numerator we get 0, in the denominator 0×0, the uncertainty is 0/0. No answer, but what is good of it?
It is that uncertainty that is revealed with the help of L’Hospital's rule. Start applying the rule. In the numerator we find the derivative of the numerator: here the derivative of the product will be 1×ln x + 1 – 1. In the denominator: the derivative of the product is ln x (the derivative of the first multiplier 1×ln x) plus the derivative of ln x (this is (x-1)/x) (the full result see on the video). Let's now check what happened to the uncertainty. You can remove units 1-1, ln 1 is 0, in the denominator: 0+0.
Again, the uncertainty is 0/0. There is no answer, but the uncertainty is the one that is good to use L’Hospital’s rule. Let's try again. In the numerator we get 1/x, in the denominator we get 1 / x + 1 / x^2 (for the convenience of differentiation, we write out the fraction (x-1)/x separately and get: (1 – 1/x)’ = 1/(x^2)). There is no need to rush to transform it, sometimes this desire arises when we have done something (applied L’Hospital’s rule). We need to stop every time, take a pause and watch if this is already the answer?
So let's substitute 1 for x. The numerator is 1, and the denominator is 1+1. The answer is received: 1/2. A mistake that should be avoided, of course, is often made - there is no uncertainty, but a person still applies L’Hospital’s rule. If we start applying L’Hospital’s rule without uncertainty, we will get an incorrect result. You can't apply L’Hospital’s rule if you don't need to! So every time we check whether there is uncertainty or not.
The following example (watch the video). This is something completely new: we practically have not calculated such limits. So, 1/x at point 0 tends to ∞, and tg 0 is 0.
The result is an uncertainty ∞ at the point 0. A completely unexpected entry. We considered power-exponential functions, but the uncertainty that was there is 1^∞ - the second special limit. We have not encountered such uncertainties. What should I do? Let's assume that the limit of this function is A, and to get rid of degrees, we always understand that the easiest way to do this is to use logarithms, and we will start calculating not A, but ln A - the logarithm of the limit.
Since the logarithm is a continuous function, we can introduce it under the limit sign (watch the video). Next, we use the properties of the logarithm: we take out the degree and write it before ln (watch the video). Now we are again converting : ln 1/x = - ln x, because x^(-1) (the exponent is again taken out the logarithm sign). Then we will shorten the entries a little. We receive: (watch the video). We send tg x to the denominator, and it becomes ctg x in the denominator. Let's now check what happens: ln at point 0 - the limit is ∞ and ctg at point 0-the limit is also ∞.
So, this is the uncertainty that is revealed using L’Hospital’s rule: - lim with x tending to 0, the numerator is the derivative of ln is 1/x, the denominator is the derivative of the cotangent 1/sin^2 x. We'll take the minus sign out of the limit again and convert it to: (watch the video). Now I will transform it like this: I will remove the square from sin and simply multiply it by sin x. It is easier to use the techniques that we have previously studied. It's just too obvious that this is where the first special limit appears. The limit of the fraction (sin x)/x is equal to 1, leaving sin x. The limit is 0.
Return back. What did we calculate? We calculated ln A=0. This means that A=1. Our goal is to find number A. So, this limit is equal to 1.
A completely new situation for us, which we would not have been able to cope with before without L’Hospital’s rule. So, and here are situations that may arise when using L’Hospital’s rule. Generally speaking, here the uncertainty ∞/∞ is at any infinity: at plus and at minus. Look, why this is so? Let's try to apply L’Hospital’s rule (watch the video). We got the same uncertainty again, but it doesn't seem to frighten us.
It happened many times that we are applying L’Hospital's rule repeatedly. We calculate the limit again by applying L’Hospital’s rule (watch the video). This is something new. Look, we've reached the same limit we started with. If we again differentiate with respect to the numerator and denominator, we’ll get the previous result. That is, in this record, these two limits will alternate infinitely many times and we will not get an answer. It's just that L’Hospital’s rule seems to work, but there is no answer to what it is equal to.
Therefore, you need to be prepared for this, too. Let's watch what can be done. It is best to transform a little: we multiply the numerator and denominator by e^x. Let's see what happens then: the numerator will have e^2x+1, and the denominator will have e^2x – 1. Nothing happened to the uncertainty - it remained, but L’Hospital’s rule will lead us to a different situation. We apply the L’Hospital’s rule: the numerator will be 2e^2x + 0, and the denominator will be 2e^2x-0 (watch the video). Look, everything is simplified here, and the answer is 1. So make an important conclusion. We can't say that L’Hospital’s rule is a lifesaver that will always save us. We must not forget about identical transformations. We must not forget about the techniques that we have studied with you in previous sections: the first and second special limits and their consequences. Well, about all the techniques that we have already discussed.