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