Use of Infinitesimals
The function α (x) is called infinitely small or an infinitesimal as x → a if
Let α (x) and β (x) be two infinitely small functions as x → a.
- If \(\lim\limits_{x \to a} {\frac{{\alpha \left( x \right)}}{{\beta \left( x \right)}}} = 0,\) then we say that the function α (x) is an infinitesimal of higher order than β (x);
- If \(\lim\limits_{x \to a} {\frac{{\alpha \left( x \right)}}{{\beta \left( x \right)}}} = A \ne 0,\) then the functions α (x) and β (x) 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 α (x) is called an infinitesimal of order \(n\) compared with the function β (x);
- If \(\lim\limits_{x \to a} {\frac{{\alpha \left( x \right)}}{{\beta \left( x \right)}}} = 1,\) then the functions α (x) and β (x) are said to be equivalent as x → 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
Example 1.
Find the limit \[\lim\limits_{x \to 0} {\frac{{\ln \left( {1 + 4x} \right)}}{{\sin 3x}}}.\]
Solution.
We use the formulas:
Then
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
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,
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