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# Calculus

Limits and Continuity of Functions

# Use of Infinitesimals

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Problems 1-2
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Problems 3-10

The function $$\alpha \left( x \right)$$ is called infinitely small or an infinitesimal as $$x \to a$$ if
$\mathop {\lim }\limits_{x \to a} \alpha \left( x \right) = 0.$ Let $$\alpha \left( x \right)$$ and $$\beta \left( x \right)$$ be two infinitely small functions as $$x \to a$$.

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

Figure 1.

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 on problem description to see solution.

### ✓Example 1

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

### ✓Example 2

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

### ✓Example 3

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

### ✓Example 4

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

### ✓Example 5

Calculate the limit $$\lim\limits_{x \to e} {\large\frac{{\ln \left( {\ln x} \right)}}{{x – e}}\normalsize}$$.

### ✓Example 6

Calculate the limit $$\lim\limits_{x \to \pi } {\large\frac{{1 + \cos x}}{{{{\left( {x – \pi } \right)}^2}}}\normalsize}$$.

### ✓Example 7

Calculate the limit $$\lim\limits_{x \to 2} {\large\frac{{{{\log }_2}x – 1}}{{x – 2}}\normalsize}$$.

### ✓Example 8

Calculate the limit $$\lim\limits_{x \to 1} {\large\frac{{\sin \left( {x – 1} \right)}}{{{x^4} – 1}}\normalsize}$$.

### ✓Example 9

Calculate the limit $$\lim\limits_{x \to 0} {\large\frac{{\ln \cos x}}{{\sqrt[3]{{1 + {x^2}}} – 1}}\normalsize}.$$

### ✓Example 10

Calculate the limit $$\lim\limits_{t \to a} {\left( {\large\frac{{\sin t}}{{\sin a}}\normalsize} \right)^{\large\frac{1}{{t – a}}\normalsize}}$$.

### Example 1.

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

#### 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 $$\lim\limits_{x \to 0} {\large\frac{{\sqrt[3]{{1 + x}} – 1}}{x}\normalsize}$$.

#### Solution.

As $$\sqrt[\large 3\normalsize]{{1 + x}} \sim 1 + {\large\frac{x}{3}\normalsize}$$, the limit can be written as
${\lim\limits_{x \to 0} \frac{{\sqrt[\large 3\normalsize]{{1 + x}} – 1}}{x} } = {\lim\limits_{x \to 0} \frac{{{{\left( {1 + x} \right)}^{\large\frac{1}{3}\normalsize}} – 1}}{x} } = {\lim\limits_{x \to 0} \frac{{1 + \frac{x}{3} – 1}}{x} } = {\frac{1}{3}\mathop {\lim }\limits_{x \to 0} \frac{x}{x} = \frac{1}{3}.}$

Page 1
Problems 1-2
Page 2
Problems 3-10