Use of Infinitesimals

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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

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