# Use of Infinitesimals

The function $$\alpha \left( x \right)$$ is called infinitely small or an infinitesimal as $$x \to a$$ if

$\lim\limits_{x \to a} \alpha \left( x \right) = 0.$

Let $$\alpha \left( x \right)$$ and $$\beta \left( x \right)$$ be two infinitely small functions as $$x \to a.$$

• If $$\lim\limits_{x \to a} {\frac{{\alpha \left( x \right)}}{{\beta \left( x \right)}}} = 0,$$ then we say that the function $$\alpha \left( x \right)$$ is an infinitesimal of higher order than $$\beta \left( x \right);$$
• If $$\lim\limits_{x \to a} {\frac{{\alpha \left( x \right)}}{{\beta \left( x \right)}}} = A \ne 0,$$ then the functions $$\alpha \left( x \right)$$ and $$\beta \left( x \right)$$ are called infinitesimals of the same order;
• If $$\lim\limits_{x \to a} {\frac{{\alpha \left( x \right)}}{{{\beta ^n}\left( x \right)}}} = A \ne 0,$$ then the function $$\alpha \left( x \right)$$ is called an infinitesimal of order $$n$$ compared with the function $$\beta \left( x \right);$$
• If $$\lim\limits_{x \to a} {\frac{{\alpha \left( x \right)}}{{\beta \left( x \right)}}} = 1,$$ then the functions $$\alpha \left( x \right)$$ and $$\beta \left( x \right)$$ are said to be equivalent as $$x \to a.$$

In particular, the following functions are equivalent:

When calculating the limit of a ratio of two infinitesimals, we can replace the terms of the ratio by their equivalent values.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the limit $\lim\limits_{x \to 0} {\frac{{\ln \left( {1 + 4x} \right)}}{{\sin 3x}}}.$

### Example 2

Find the limit $\lim\limits_{x \to 0} {\frac{{\sqrt[3]{{1 + x}} - 1}}{x}}.$

### Example 3

Find the limit $\lim\limits_{t \to 0} {\frac{{1 - \cos \left( {1 - \cos t} \right)}}{{{{\sin }^2}{t^2}}}}.$

### Example 4

Calculate the limit $\lim\limits_{x \to 0} {\frac{{\sqrt {1 + 2x + 3{x^2}} - 1}}{x}}.$

### Example 1.

Find the limit $\lim\limits_{x \to 0} {\frac{{\ln \left( {1 + 4x} \right)}}{{\sin 3x}}}.$

Solution.

We use the formulas:

$\ln \left( {1 + \alpha } \right) \sim \alpha ,\;\;\sin \alpha \sim \alpha .$

Then

$\lim\limits_{x \to 0} \frac{{\ln \left( {1 + 4x} \right)}}{{\sin 3x}} = \lim\limits_{x \to 0} \frac{{4x}}{{3x}} = \frac{4}{3}.$

### Example 2.

Find the limit $\lim\limits_{x \to 0} {\frac{{\sqrt[3]{{1 + x}} - 1}}{x}}.$

Solution.

As $$\sqrt[3]{{1 + x}} \sim 1 + {\frac{x}{3}},$$ the limit can be written as

$\lim\limits_{x \to 0} \frac{{\sqrt[3]{{1 + x}} - 1}}{x} = \lim\limits_{x \to 0} \frac{{{{\left( {1 + x} \right)}^{\frac{1}{3}}} - 1}}{x} = \lim\limits_{x \to 0} \frac{{1 + \frac{x}{3} - 1}}{x} = \frac{1}{3}\lim\limits_{x \to 0} \frac{\cancel{x}}{\cancel{x}} = \frac{1}{3}.$

### Example 3.

Find the limit $\lim\limits_{t \to 0} {\frac{{1 - \cos \left( {1 - \cos t} \right)}}{{{{\sin }^2}{t^2}}}}.$

Solution.

We know that $$\cos t \sim 1 - {\frac{{{t^2}}}{2}}$$ and $$\sin t \sim t$$ as $$t \to 0.$$ Hence,

$\lim\limits_{t \to 0} \frac{{1 - \cos \left( {1 - \cos t} \right)}}{{{{\sin }^2}{t^2}}} = \lim\limits_{t \to 0} \frac{{1 - \cos \left( {1 - 1 + \frac{{{t^2}}}{2}} \right)}}{{{{\left( {\sin {t^2}} \right)}^2}}} = \lim\limits_{t \to 0} \frac{{1 - \cos \frac{{{t^2}}}{2}}}{{{{\left( {{t^2}} \right)}^2}}} = \lim\limits_{t \to 0} \frac{{1 - \left[ {1 - \frac{1}{2}{{\left( {\frac{{{t^2}}}{2}} \right)}^2}} \right]}}{{{t^4}}} = \lim\limits_{t \to 0} \frac{{\frac{{{t^4}}}{8}}}{{{t^4}}} = \frac{1}{8}.$

### Example 4.

Calculate the limit $\lim\limits_{x \to 0} {\frac{{\sqrt {1 + 2x + 3{x^2}} - 1}}{x}}.$

Solution.

Replacing the square root with the equivalent infinitely small function, we have

$\lim\limits_{x \to 0} \frac{{\sqrt {1 + 2x + 3{x^2}} - 1}}{x} = \lim\limits_{x \to 0} \frac{{1 + \frac{{2x + 3{x^2}}}{2} - 1}}{x} = \frac{1}{2}\lim\limits_{x \to 0} \frac{{2x + 3{x^2}}}{x} = \frac{1}{2}\lim\limits_{x \to 0} \left( {2 + 3x} \right) = \frac{1}{2} \cdot 2 = 1.$