The first and second wonderful limits

In this topic we will talk about wonderful limits. There are several wonderful limits, but the most famous are the first and second wonderful limits. The remarkable thing about these limits is that they have a wide application and can be used to find other limits found in numerous problems. This is what we will do in the practical part of this lesson. To solve problems by reducing them to the first or second wonderful limit, it is not necessary to disclose the uncertainties they contain, since the values of these limits have long been deduced by great mathematicians.

The first wonderful limit

Let's consider the following limit:

$$\lim\limits_{x\to0} \dfrac{\sin x}{x}$$According to the rule for finding limits, we try to substitute zero into the function: in the numerator we get zero (the sine of zero is zero), in the denominator, obviously, we also get zero. Thus, we have a $\dfrac{0}{0}$ uncertainty. We do not need to reveal it. In the course of mathematical analysis, it is proved that

$$\lim\limits_{x\to0} \dfrac{\sin x}{x}=1$$This fact is called the first wonderful limit.

The functions can be arranged differently, this does not change anything:

$$\lim\limits_{x\to0} \dfrac{x}{\sin x}=1$$In this expression, the variable is $x$. In practice, not only a variable can act as a parameter, but also an elementary function or a complex function. It is only important that this function should tend to $0$. For example:

$$\lim\limits_{x\to0} \dfrac{\sin 8x}{8x} =1$$$$\lim\limits_{x\to0} \dfrac{\frac{2x}{3}}{\sin \frac{2x}{3}}=1$$$$\lim\limits_{x \to 0} \dfrac{\sin (2x^2 + 4x + x)}{(2x^2 + 4x + x) }=1$$In these examples $8x \to 0$, $\dfrac{2x}{3}\to0$ and $2x^2 + 4x + x\to 0$, when $x \to 0$. So, we could write that these limits are equal to $1$.

Examples of use of the first wonderful limit

Let's find the following limit:

$$\lim\limits_{x\to0} \dfrac{\sin8x}{5x}$$First, we try to substitute $0$ in the expression under the limit sign. We will get a $\dfrac{0}{0}$ uncertainty. We see that the expression under the limit sign looks like the first wonderful limit, but this is not exactly it: under the sine there is $8x$, and in the denominator there is $5x$. In such cases, we need to organize the first wonderful limit ourselves, using the following simple method:

$$\lim\limits_{x\to0} \dfrac{\sin8x}{5x} = \lim\limits_{x\to0} \dfrac{\sin8x}{5 \cdot \dfrac{1}{8}\cdot8x} = \lim\limits_{x\to0} \dfrac{\sin8x}{ \dfrac{5}{8}\cdot8x} =\dfrac{1}{\dfrac{5}{8}} \lim\limits_{x\to0} \dfrac{\sin8x}{8x} = \dfrac{8}{5} \underbrace{\lim\limits_{x\to0} \dfrac{\sin8x}{8x}}_{=1} = \dfrac{8}{5}$$Now let's find another limit:

$$\lim\limits_{x\to0} \dfrac{3x^2}{\sin^2\dfrac{x}{4}}$$Yet again, we try to substitute 0 in the expression under the limit sign. We will get a $\dfrac{0}{0}$ uncertainty. Let's factorize the numerator and the denominator:

$$\lim\limits_{x\to0} \dfrac{3x^2}{\sin^2\dfrac{x}{4}} = \lim\limits_{x\to0} \dfrac{3x \cdot x}{\sin \dfrac{x}{4} \cdot \sin \dfrac{x}{4}}$$Now again we organize the first wonderful limits. Under the sines we have $\dfrac{x}{4}$, so we also need to get $\dfrac{x}{4}$ in the numerator:

$$\lim\limits_{x\to0} \dfrac{3x^2}{\sin^2\dfrac{x}{4}} = \lim\limits_{x\to0} \dfrac{3x \cdot x}{\sin \dfrac{x}{4} \cdot \sin \dfrac{x}{4}} = \lim\limits_{x\to0} \dfrac{3 \cdot 4 \cdot 4 \cdot \dfrac{x}{4} \cdot \dfrac{x}{4}}{\sin \dfrac{x}{4} \cdot \sin \dfrac{x}{4}} = 3 \cdot 4 \cdot 4 = 48$$Here we have used the fact that $$\lim\limits_{x \to0} \dfrac{\dfrac{x}{4}}{\sin \dfrac{x}{4}}=1$$

The second wonderful limit

Let's consider the following limit:

$$\lim\limits_{x \to \infty} \left( 1 + \dfrac{1}{x}\right)^{x}$$If we substitute the infinity for $x$, we get 1∞ uncertainty. In the course of mathematical analysis, it is proven that

$$\lim\limits_{x \to \infty} \left( 1 + \dfrac{1}{x}\right)^{x} = e$$This fact is called the second wonderful limit.

The second wonderful limit can be written in another form:

$$\lim\limits_{x \to 0} \left( 1 + x\right)^{\frac{1}{x}} = e$$The parameter $x$ can be not only a variable, but also a complex function. It is only important that it must tend to infinity. For example:

$$\lim\limits_{x \to \infty} \left( 1 + \dfrac{1}{5x}\right)^{7x}$$We notice that $7x \to \infty$ and $1 + \dfrac{1}{5x} \to 1$, if $x \to \infty$. We have $1^{\infty}$ uncertainty. This uncertainty is just revealed with the second wonderful limit. In this example in the denominator we have $5x$, so, in the exponent, we also need to have $5x$. For this, we raise the base to the power of $5x$, and so that the expression does not change, we raise it to the power of $\dfrac{1}{5x}$:

$$\lim\limits_{x \to \infty} \left( 1 + \dfrac{1}{5x}\right)^{7x} = \lim\limits_{x \to \infty} \left(\left( 1 + \dfrac{1}{5x}\right)^{5x}\right)^{\frac{1}{5x} \cdot 7x} = e^{\lim\limits_{x\to \infty} \frac{7x}{5x} }= e^{\frac{7}{5}}$$because $$\lim\limits_{x \to \infty}\left(1 + \dfrac{1}{5x}\right)^{5x} = e$$

Examples of use of the second wonderful limit

Let's find the following limit:

$$\lim\limits_{x \to \infty} \left( \dfrac{x+5}{x-2}\right)^{3x+1}$$If we substitute the infinity for $x$, we get the $\left(\dfrac{\infty}{\infty}\right)^\infty$ uncertainty. Infinity in the exponent is a sign that the expression can be reduced to the ratio of two second wonderful limits. Indeed, if we divided the numerator and denominator by $x$, we will get

$$\lim\limits_{x \to \infty} \left( \dfrac{x+5}{x-2}\right)^{3x+1} = \lim\limits_{x \to \infty} \left( \dfrac{\frac{x}{x} + \frac{5}{x}}{\frac{x}{x} - \frac{2}{x}} \right)^{3x+1} = \lim\limits_{x \to \infty} \left( \dfrac{1 + \frac{5}{x}}{1 - \frac{2}{x}} \right)^{3x+1}$$To take advantage of the second wonderful limit, it is necessary that the second terms in both the numerator and the denominator are equal to 1. To do this, we will replace the functions:

$$\dfrac{5}{x} = \dfrac{1}{n} \ \Rightarrow \ x =5n \ \Rightarrow \ 3x+1 = 15n+1 \\ \ \\ -\dfrac{2}{x} = \dfrac{1}{t} \ \Rightarrow \ x = -2t \ \Rightarrow \ 3x+1 = -6t+1$$Substituting this into our expression and continuing the equality, we get:

$$\lim\limits_{x \to \infty} \left( \dfrac{1 + \frac{5}{x}}{1 - \frac{2}{x}} \right)^{3x+1} = \dfrac{\lim\limits_{n \to \infty} \left(1 + \frac{1}{n}\right)^{15n+1}}{{\lim\limits_{t \to \infty} \left(1 + \frac{1}{t}\right)^{-6t+1}}}$$This is already the ratio of the second wonderful limits:

$$\dfrac{\lim\limits_{n \to \infty} \left(1 + \frac{1}{n}\right)^{15n+1}}{{\lim\limits_{t \to \infty} \left(1 + \frac{1}{t}\right)^{-6t+1}}} = \dfrac{e^{15}}{e^{-6}} = e^{15+6} = e^{21}$$

Conclusion

In this topic, we have learned about the first and second wonderful limits. These limits help to greatly simplify the work of finding the limits. But in order to be able to bring your limit to the wonderful limit, you need to take a good look at it, because they do not occur in a direct form, but often in the form of consequences, equipped with additional terms and factors.