# 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

$L = \lim\limits_{y \to 0} {\left( {1 - \frac{{{y^2}}}{2} + y\cot a} \right)^{\frac{1}{y}}} = \lim\limits_{y \to 0} {\left( {1 + y\cot a} \right)^{\frac{1}{y}}} = \lim\limits_{y \to 0} {\left( {1 + y\cot a} \right)^{\frac{{\cot a}}{{y\cot a}}}} = \left[ {\lim\limits_{y\cot a \to 0} {{\left( {1 + y\cot a} \right)}^{\frac{1}{{y\cot a}}}}} \right]^{\cot a} = e^{\cot a}.$