Use of Infinitesimals
Solved Problems
Example 5.
Calculate the limit \[\lim\limits_{x \to e} {\frac{{\ln \left( {\ln x} \right)}}{{x - e}}}.\]
Solution.
We apply the formula \(\ln \left( {1 + \alpha } \right) \sim \alpha \) as \(\alpha \to 0.\) This yields:
\[\lim\limits_{x \to e} \frac{{\ln \left( {\ln x} \right)}}{{x - e}} = \lim\limits_{x \to e} \frac{{\ln \left( {\ln x + 1 - 1} \right)}}{{x - e}} = \lim\limits_{x \to e} \frac{{\ln \left[ {1 + \left( {\ln x - 1} \right)} \right]}}{{x - e}} = \lim\limits_{x \to e} \frac{{\ln x - 1}}{{x - e}} = \lim\limits_{x \to e} \frac{{\ln x - \ln e}}{{x - e}} = \lim\limits_{x \to e} \frac{{\ln \frac{x}{e}}}{{x - e}} = \lim\limits_{x \to e} \frac{{\ln \left[ {1 + \left( {\frac{x}{e} - 1} \right)} \right]}}{{x - e}} = \lim\limits_{x \to e} \frac{{\frac{x}{e} - 1}}{{x - e}} = \lim\limits_{x \to e} \frac{{\frac{{x - e}}{e}}}{{x - e}} = \frac{1}{e}\lim\limits_{x \to e} \frac{\cancel{x - e}}{\cancel{x - e}} = \frac{1}{e}.\]
Example 6.
Calculate the limit \[\lim\limits_{x \to \pi } {\frac{{1 + \cos x}}{{{{\left( {x - \pi } \right)}^2}}}}.\]
Solution.
Change the variable: \(x -\pi = y.\) Here \(y \to 0\) as \(x \to \pi.\) Hence,
\[L = \lim\limits_{x \to \pi } \frac{{1 + \cos x}}{{{{\left( {x - \pi } \right)}^2}}} = \lim\limits_{y \to 0} \frac{{1 + \cos \left( {y + \pi } \right)}}{{{y^2}}}.\]
Using the reduction formula \(\cos \left( {y + \pi } \right) = - \cos y,\) we obtain
\[L = \lim\limits_{y \to 0} \frac{{1 - \cos y}}{{{y^2}}}.\]
Finally, replacing the cosine with the equivalent infinitesimal \(1 - \cos y \sim {\frac{{{y^2}}}{2}},\) we find the limit:
\[L = \lim\limits_{y \to 0} \frac{{1 - \cos y}}{{{y^2}}} = \lim\limits_{y \to 0} \frac{{\frac{{{y^2}}}{2}}}{{{y^2}}} = \frac{1}{2}.\]
Example 7.
Calculate the limit \[\lim\limits_{x \to 2} {\frac{{{{\log }_2}x - 1}}{{x - 2}}}.\]
Solution.
Using the equivalent expression for logarithm: \(\ln \left( {1 + \alpha } \right) \sim \alpha \) as \(\alpha \to 0,\) we have
\[\lim\limits_{x \to 2} \frac{{{{\log }_2}x – 1}}{{x – 2}} = \lim\limits_{x \to 2} \frac{{{{\log }_2}x – {{\log }_2}2}}{{x – 2}} = \lim\limits_{x \to 2} \frac{{{{\log }_2}\frac{x}{2}}}{{x – 2}} = \lim\limits_{x \to 2} \frac{{\frac{{\ln \left( {x/2} \right)}}{{\ln 2}}}}{{x – 2}} = \frac{1}{{\ln 2}}\lim\limits_{x \to 2} \frac{{\ln \left( {x/2} \right)}}{{x – 2}} = \frac{1}{{\ln 2}}\lim\limits_{x \to 2} \frac{{\ln \left[ {1 + \left( {\frac{x}{2} – 1} \right)} \right]}}{{x – 2}} = \frac{1}{{\ln 2}}\lim\limits_{x \to 2} \frac{{\frac{x}{2} – 1}}{{x – 2}} = \frac{1}{{2\ln 2}}\lim\limits_{x \to 2} \frac{\cancel{x – 2}}{\cancel{x – 2}} = \frac{1}{{2\ln 2}}.\]
Example 8.
Calculate the limit \[\lim\limits_{x \to 1} {\frac{{\sin \left( {x - 1} \right)}}{{{x^4} - 1}}}.\]
Solution.
Let \(x -1 = t.\) Then \(t \to 0\) as \(x \to 1,\) so that the limit becomes
\[L =\lim\limits_{x \to 1} \frac{{\sin \left( {x - 1} \right)}}{{{x^4} - 1}} = \lim \limits_{t \to 0} \frac{{\sin t}}{{{{\left( {t + 1} \right)}^4} - 1}}.\]
Now we use the algebraic identity
\[\left( {t + 1} \right)^4 = {t^4} + 4{t^3} + 6{t^2} + 4t + 1\]
to find the limit:
\[L = \lim\limits_{t \to 0} \frac{{\sin t}}{{{{\left( {t + 1} \right)}^4} - 1}} = \lim\limits_{t \to 0} \frac{{\sin t}}{{\left( {{t^4} + 4{t^3} + 6{t^2} + 4t + 1} \right) - 1}} = \left[ {\sin t \sim t} \right] = \lim\limits_{t \to 0} \frac{\cancel{t}}{{\cancel{t}\left( {{t^3} + 4{t^2} + 6t + 4} \right)}} = \lim\limits_{t \to 0} \frac{1}{{{t^3} + 4{t^2} + 6t + 4}} = \frac{1}{4}.\]
Example 9.
Calculate the limit \[\lim\limits_{x \to 0} {\frac{{\ln \cos x}}{{\sqrt[3]{{1 + {x^2}}} - 1}}}.\]
Solution.
We use the following equivalent representations:
\[\sqrt[k]{{1 + \alpha }} \sim 1 + \frac{\alpha }{k},\;\;\; \ln \left( {1 + \alpha } \right) \sim \alpha \;\;\text{as}\;\;\alpha \to 0.\]
Then
\[L = \lim\limits_{x \to 0} \frac{{\ln \cos x}}{{\sqrt[3]{{1 + {x^2}}} - 1}} = \lim\limits_{x \to 0} \frac{{\ln \left[ {1 + \left( {\cos x - 1} \right)} \right]}}{{\left( {1 + \frac{{{x^2}}}{3}} \right) - 1}} = \lim\limits_{x \to 0} \frac{{\cos x - 1}}{{{x^2}/3}} = - 3\lim\limits_{x \to 0} \frac{{1 - \cos x}}{{{x^2}}}.\]
Replacing \(1 - \cos x \sim {\frac{{{x^2}}}{2}},\) we get the final answer:
\[L = - 3\lim\limits_{x \to 0} \frac{{1 - \cos x}}{{{x^2}}} = - 3\lim\limits_{x \to 0} \frac{{{x^2}/2}}{{{x^2}}} = - \frac{3}{2}\lim\limits_{x \to 0} \frac{{\cancel{x^2}}}{{\cancel{x^2}}} = - \frac{3}{2}.\]
Example 10.
Calculate the limit \[\lim\limits_{t \to a} {\left( {\frac{{\sin t}}{{\sin a}}} \right)^{\frac{1}{{t - a}}}}.\]
Solution.
Change the variable: \(t - a = y\;\; ,\) \( \Rightarrow y \to 0\) as \(t \to a.\) Then the limit can be written in terms of the new variable \(y\) as
\[L = \lim\limits_{t \to a} {\left( {\frac{{\sin t}}{{\sin a}}} \right)^{\frac{1}{{t - a}}}} = \lim\limits_{y \to 0} {\left( {\frac{{\sin \left( {y + a} \right)}}{{\sin a}}} \right)^{\frac{1}{y}}} = \lim\limits_{y \to 0} {\left( {\frac{{\sin y\cos a + \cos y\sin a}}{{\sin a}}} \right)^{\frac{1}{y}}} = \lim\limits_{y \to 0} {\left( {\cos y + \sin y\cot a} \right)^{\frac{1}{y}}}.\]
Replace the functions of cosine and sine with their equivalent infinitely small functions: \(\cos y \sim 1 - {\frac{{{y^2}}}{2}},\) \(\sin y \sim y.\) Then the limit becomes
\[L = \lim\limits_{y \to 0} {\left( {\cos y + \sin y\cot a} \right)^{\frac{1}{y}}} = \lim\limits_{y \to 0} {\left( {1 - \frac{{{y^2}}}{2} + y\cot a} \right)^{\frac{1}{y}}}.\]
We can consider only first-order infinitesimals and neglect second-order infinitesimals \({\frac{{{y^2}}}{2}}.\) As a result, we obtain the final answer
\[L = \lim\limits_{y \to 0} {\left( {1 - \frac{{{y^2}}}{2} + y\cot a} \right)^{\frac{1}{y}}} = \lim\limits_{y \to 0} {\left( {1 + y\cot a} \right)^{\frac{1}{y}}} = \lim\limits_{y \to 0} {\left( {1 + y\cot a} \right)^{\frac{{\cot a}}{{y\cot a}}}} = \left[ {\lim\limits_{y\cot a \to 0} {{\left( {1 + y\cot a} \right)}^{\frac{1}{{y\cot a}}}}} \right]^{\cot a} = e^{\cot a}.\]
