Calculus

Applications of the Derivative

Applications of Derivative Logo

Asymptotes

Solved Problems

Example 15.

Find the asymptotes of the function \[f\left( x \right) = x + \sqrt {{x^2} - 1} .\]

Solution.

Examine the domain of the function:

\[{x^2} - 1 \ge 0,\;\; \Rightarrow {x^2} \ge 1,\;\; \Rightarrow x \in \left( { - \infty , - 1} \right] \cup \left[ {1, + \infty } \right).\]

Check for vertical asymptotes at the boundary points \(x = \pm 1:\)

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

We see that the function has no vertical asymptotes.

Now we look for horizontal asymptotes:

\[\lim\limits_{x \to + \infty } f\left( x \right) = \lim\limits_{x \to + \infty } \left( {x + \sqrt {{x^2} - 1} } \right) = \infty ;\]
\[\lim\limits_{x \to - \infty } f\left( x \right) = \lim\limits_{x \to - \infty } \left( {x + \sqrt {{x^2} - 1} } \right) = \left[ {\infty - \infty } \right].\]

In the last limit, there is an indeterminate form of kind \(\infty - \infty.\) To solve this limit we rationalize the expression:

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

So we get a horizontal asymptote at \(y = 0\) in the left direction (as \(x \to -\infty).\)

Consider a possible oblique asymptote as \(x \to +\infty:\)

\[k = \lim\limits_{x \to + \infty } \frac{{f\left( x \right)}}{x} = \lim\limits_{x \to + \infty } \frac{{x + \sqrt {{x^2} - 1} }}{x} = \lim\limits_{x \to + \infty } \left( {1 + \sqrt {\frac{{{x^2} - 1}}{{{x^2}}}} } \right) = 1 + \lim\limits_{x \to + \infty } \sqrt {1 - \frac{1}{{{x^2}}}} = 1 + 1 = 2;\]
\[b = \lim\limits_{x \to + \infty } \left[ {f\left( x \right) - kx} \right] = \lim\limits_{x \to + \infty } \left( {x + \sqrt {{x^2} - 1} - 2x} \right) = \lim\limits_{x \to + \infty } \left( {\sqrt {{x^2} - 1} - x} \right) = \left[ {\infty - \infty } \right].\]

We again rationalize this expression:

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

The oblique asymptote is given by the equation \(y =2x\) as \(x \to +\infty.\)

Thus, the function has two asymptotes: \(y = 0\) (as \(x \to -\infty)\) and \(y = 2x\) (as \(x \to \infty).\)

Asymptotes of the function f(x)=x+sqrt(x^2-1)
Figure 17.

Example 16.

Find the asymptotes of the function \[f\left( x \right) = \sqrt[3]{{{x^3} - {x^2}}}.\]

Solution.

The function is defined for all \(x \in \mathbb{R}\), so it has no vertical asymptotes.

Examine possible horizontal asymptotes.

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

Similarly for \(x \to -\infty.\) Hence, there are no horizontal asymptotes.

Look now for oblique asymptotes. We calculate both limits \(x \to \pm\infty:\)

\[k = \lim\limits_{x \to + \infty } \frac{{f\left( x \right)}}{x} = \lim\limits_{x \to + \infty } \frac{{\sqrt[3]{{{x^3} - {x^2}}}}}{x} = \lim\limits_{x \to + \infty } \sqrt[3]{{\frac{{{x^3} - {x^2}}}{{{x^3}}}}} = \lim\limits_{x \to + \infty } \sqrt[3]{{1 - \frac{1}{x}}} = 1;\]
\[b = \lim\limits_{x \to + \infty } \left[ {f\left( x \right) - kx} \right] = \lim\limits_{x \to + \infty } \left[ {\sqrt[3]{{{x^3} - {x^2}}} - x} \right] = \left[ {\infty - \infty } \right].\]

We rationalize the latter expression using the identity:

\[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right).\]

Consequently,

\[b = \lim\limits_{x \to + \infty } \left[ {\sqrt[3]{{{x^3} - {x^2}}} - x} \right] = \lim\limits_{x \to + \infty } \frac{{{{\left( {\sqrt[3]{{{x^3} - {x^2}}}} \right)}^3} - {x^3}}}{{{{\left( {\sqrt[3]{{{x^3} - {x^2}}}} \right)}^2} + x\sqrt[3]{{{x^3} - {x^2}}} + {x^2}}} = \lim\limits_{x \to + \infty } \frac{{\cancel{x^3} - {x^2} - \cancel{x^3}}}{{{{\left( {\sqrt[3]{{{x^3} - {x^2}}}} \right)}^2} + x\sqrt[3]{{{x^3} - {x^2}}} + {x^2}}} = - \lim\limits_{x \to + \infty } \frac{{{x^2}}}{{\sqrt[3]{{{{\left( {{x^3} - {x^2}} \right)}^2}}} + x\sqrt[3]{{{x^3} - {x^2}}} + {x^2}}} = - \lim\limits_{x \to + \infty } \frac{{{x^2}}}{{\sqrt[3]{{{x^6} - 2{x^5} + {x^4}}} + \sqrt[3]{{{x^6} - {x^5}}} + {x^2}}} = - \lim\limits_{x \to + \infty } \frac{1}{{\sqrt[3]{{1 - \frac{2}{x} + \frac{1}{{{x^2}}}}} + \sqrt[3]{{1 - \frac{1}{x}}} + 1}} = - \frac{1}{3}.\]

So the oblique asymptote is given by the equation \(y = x - \frac{1}{3}.\)

Asymptotes of the function f(x)=(x^3-x^2)^(1/3)
Figure 18.

Example 17.

Find the asymptotes of the curve defined by the parametric equations

\[x = \varphi \left( t \right) = \frac{1}{{t - 2}},y = \psi \left( t \right) = \frac{t}{{t - 1}}.\]

Solution.

Examine the point \(t = 2.\) Calculate the limits of the functions \(x = \varphi \left( t \right)\) and \(y = \psi \left( t \right)\) as \(t \to 2:\)

\[\lim\limits_{t \to 2} \varphi \left( t \right) = \lim\limits_{t \to 2} \frac{1}{{t - 2}} = \infty ,\;\;{\lim\limits_{t \to 2} \psi \left( t \right) = \lim\limits_{t \to 2} \frac{t}{{t - 1}} = 2.}\]

Thus, the \(x\)-coordinate tends to infinity and the \(y\)-coordinate approaches the finite value \(y = 2\) as \(t \to 2.\) Consequently, the curve has the horizontal asymptote \(y = 2.\)

Similarly, consider the point \(t = 1,\) where the function \(y\left( t \right)\) has a discontinuity:

\[\lim\limits_{t \to 1} \varphi \left( t \right) = \lim\limits_{t \to 1} \frac{1}{{t - 2}} = -1,\;\;\lim\limits_{t \to 1} \psi \left( t \right) = \lim\limits_{t \to 1} \frac{t}{{t - 1}} = \infty.\]

Hence, \(x = -1\) is a vertical asymptote of the curve.

It follows from the form of the function \(x = \varphi \left( t \right)\) that \(t = 2\) is the single value of the parameter \(t\) at which the \(x\)-coordinate tends to infinity. Hence, the curve has no oblique asymptotes.

Thus, this curve the horizontal asymptote \(y = 2\) and the vertical asymptote \(x = -1.\) Its schematic view is shown in Figure \(19.\)

Asymptotes of a parametric function
Figure 19.

Example 18.

Find the asymptotes of the curve defined by the parametric equations

\[x = \varphi \left( t \right) = \frac{t}{{1 - {t^2}}}, y = \psi \left( t \right) = \frac{{{t^2}}}{{1 - {t^2}}}.\]

Solution.

When \(t = \pm 1,\) the denominators in the expressions for \(x\) and \(y\) are equal to zero. Therefore, we have infinite limits at these value of \(t:\)

\[\lim\limits_{t \to \pm 1} \varphi \left( t \right) = \lim\limits_{t \to \pm 1} \frac{t}{{1 - {t^2}}} = \infty ,\;\;\lim\limits_{t \to \pm 1} \psi \left( t \right) = \lim\limits_{t \to \pm 1} \frac{{{t^2}}}{{1 - {t^2}}} = \infty .\]

Consequently, oblique asymptotes may exist at \(t = -1\) and \(t = 1\). We investigate first the point \(t = -1:\)

\[{k_1} = \lim\limits_{t \to - 1} \frac{{\psi \left( t \right)}}{{\varphi \left( t \right)}} = \lim\limits_{t \to - 1} \frac{{\frac{{{t^2}}}{{1 - {t^2}}}}}{{\frac{t}{{1 - {t^2}}}}} = \lim\limits_{t \to - 1} \frac{{{t^2}}}{t} = \lim\limits_{t \to - 1} t = - 1,\]
\[{b_1} = \lim\limits_{t \to - 1} \left[ {\psi \left( t \right) - {k_1}\varphi \left( t \right)} \right] = \lim\limits_{t \to - 1} \left[ {\frac{{{t^2}}}{{1 - {t^2}}} - \left( { - 1} \right)\frac{t}{{1 - {t^2}}}} \right] = \lim\limits_{t \to - 1} \frac{{{t^2} + t}}{{1 - {t^2}}} = \lim\limits_{t \to - 1} \frac{{t\cancel{\left( {t + 1} \right)}}}{{\left( {1 - t} \right)\cancel{\left( {1 + t} \right)}}} = \lim\limits_{t \to - 1} \frac{t}{{1 - t}} = - \frac{1}{2}.\]

Thus, at \(t = -1,\) there is an oblique asymptote, which is given by the equation

\[y = {k_1}x + {b_1} = - x - \frac{1}{2}.\]

Similarly, we find the equation of the asymptote at \(t = 1:\)

\[{k_2} = \lim\limits_{t \to 1} \frac{{\psi \left( t \right)}}{{\varphi \left( t \right)}} = \lim\limits_{t \to 1} \frac{{\frac{{{t^2}}}{{1 - {t^2}}}}}{{\frac{t}{{1 - {t^2}}}}} = \lim\limits_{t \to 1} \frac{{{t^2}}}{t} = \lim\limits_{t \to 1} t = 1,\]
\[{b_2} = \lim\limits_{t \to 1} \left[ {\psi \left( t \right) - {k_2}\varphi \left( t \right)} \right] = \lim\limits_{t \to 1} \left[ {\frac{{{t^2}}}{{1 - {t^2}}} - \frac{t}{{1 - {t^2}}}} \right] = \lim\limits_{t \to 1} \frac{{{t^2} - t}}{{1 - {t^2}}} = \lim\limits_{t \to 1} \frac{{t\left( {t - 1} \right)}}{{\left( {1 - t} \right)\left( {1 + t} \right)}} = - \lim\limits_{t \to 1} \frac{t}{{1 + t}} = - \frac{1}{2}.\]

So, there is one more asymptote at this point. Its equation is written as

\[y = {k_2}x + {b_2} = x - \frac{1}{2}.\]

There are no other values of \(t\) such that the coordinates \(x\) and \(y\) tend to infinities. Hence, the curve has no other asymptotes.

This curve consists of two branches. The lower branch is not defined at \(x = 0\) (when \(t \to \pm\infty\)). The shape of the curve is shown in Figure \(20.\)

Asymptotes of a curve defined by parametric equations
Figure 20.

Example 19.

Find the asymptotes of the hyperbolic spiral given by the equation \[\rho = {\frac{a}{\varphi }}.\]

Solution.

We express the angle \(\varphi\) in terms of the radius \(\rho:\)

\[\varphi = \frac{a}{\rho }.\]

An asymptote in polar coordinates is defined by the parameters \(\alpha\) and \(p\). We first calculate the angle \(\alpha:\)

\[\alpha = \lim\limits_{\rho \to \infty } \varphi = \lim\limits_{\rho \to \infty } \frac{a}{\rho } = 0,\]

that is the asymptote (if it exists) is positioned horizontally. Determine the parameter \(p:\)

\[p = \lim\limits_{\rho \to \infty } \left[ {\rho \sin \left( {\alpha - \varphi } \right)} \right] = \lim\limits_{\rho \to \infty } \left[ {\rho \sin \left( {0 - \varphi } \right)} \right] = - \lim\limits_{\rho \to \infty } \left( {\rho \sin \varphi } \right) = - \lim\limits_{\rho \to \infty } \left( {\rho \sin \frac{a}{\rho }} \right) = - \lim\limits_{\rho \to \infty } \frac{{\sin \frac{a}{\rho }}}{{\frac{1}{\rho }}} = - \lim\limits_{\frac{a}{\rho } \to 0} \frac{{a\sin \frac{a}{\rho }}}{{\frac{a}{\rho }}} = - a\lim\limits_{\frac{a}{\rho } \to 0} \frac{{\sin \frac{a}{\rho }}}{{\frac{a}{\rho }}} = - a \cdot 1 = - a.\]

Here we took into account that the special trigonometric limit is equal to \(1.\)

Hence, the hyperbolic spiral has a horizontal asymptote with the parameters \(\alpha =0\), \(p = -a.\) Its equation in Cartesian coordinates has the form \(y = a\) (Figure \(21\)).

Asymptote of the hyperbolic spiral
Figure 21.

Example 20.

Find the asymptotes of the curve given in polar coordinates: \[\rho = a\tan \varphi .\]

Solution.

First we express the inverse relationship \(\varphi \left( \rho \right)\) from the equation:

\[\rho = a\tan \varphi ,\;\; \Rightarrow \frac{\rho }{a} = \tan \varphi ,\;\; \Rightarrow \varphi = \arctan \frac{\rho }{a}.\]

An asymptote in polar coordinates is determined by two parameters: the angle \(\alpha\) and the distance \(p\) from the pole to the asymptote. The angle \(\alpha\) is calculated as follows:

\[\alpha = \lim\limits_{\rho \to \infty } \varphi = \lim\limits_{\rho \to \infty } \arctan \frac{\rho }{a} = \lim\limits_{\frac{\rho }{a} \to \infty } \arctan \frac{\rho }{a} = \frac{\pi }{2}.\]

The parameter \(p\) is respectively given by

\[p = \lim\limits_{\rho \to \infty } \left[ {\rho \sin \left( {\alpha - \varphi } \right)} \right] = \lim\limits_{\rho \to \infty } \left[ {\rho \sin \left( {\frac{\pi }{2} - \varphi } \right)} \right] = \lim\limits_{\rho \to \infty } \left[ {\rho \cos \varphi } \right] = \lim\limits_{\rho \to \infty } \left[ {\rho \cos \left( {\arctan \frac{\rho }{a}} \right)} \right].\]

Next, we use the following identity:

\[\arctan z = \arccos \frac{1}{{\sqrt {1 + {z^2}} }}.\]

Then

\[\rho \cos \left( {\arctan \frac{\rho }{a}} \right) = \rho \cos \left( {\arccos \frac{1}{{\sqrt {1 + {{\left( {\frac{\rho }{a}} \right)}^2}} }}} \right) = \rho \cos \left( {\arccos \sqrt {\frac{{{a^2}}}{{{a^2} + {\rho ^2}}}} } \right) = \pm \frac{{\rho a}}{{\sqrt {{a^2} + {\rho ^2}} }}.\]

Hence, the parameter \(p\) is equal to:

\[p = \lim\limits_{\rho \to \infty } \left( { \pm \frac{{\rho a}}{{\sqrt {{a^2} + {\rho ^2}} }}} \right) = \pm \lim\limits_{\rho \to \infty } \frac{a}{{\sqrt {\frac{{{a^2}}}{{{\rho ^2}}} + 1} }} = \pm a.\]

Thus, we have two values of the parameter \(p:\) \(p = +a\) and \(p = -a,\) i.e. there are two asymptotes. Since the angle \(\alpha\) is the same for both asymptotes and is equal to \(\frac{\pi }{2},\) these asymptotes are vertical. In this case, \(p\) is the distance from the center to the asymptote along the \(x\)-axis. In Cartesian coordinates, the equations of the asymptotes look as follows: \(x = a\) and \(x = -a.\) This is shown schematically in Figure \(22.\)

Asymptotes of the polar curve r=a*tan(phi)
Figure 22.

Example 21.

Find the asymptotes of the folium of Descartes given by the equation \[{x^3} + {y^3} = 3axy.\]

Solution.

We first investigate the oblique asymptotes of the curve. Write the implicit equation in the form

\[{x^3} + {y^3} - 3axy = 0.\]

Substituting the equation of the asymptote \(y =kx + b,\) we obtain:

\[{x^3} + {\left( {kx + b} \right)^3} - 3ax\left( {kx + b} \right) = 0,\;\; \Rightarrow {x^3} + {k^3}{x^3} + 3{k^2}b{x^2} + 3k{b^2}x + {b^3} - 3ak{x^2} - 3abx = 0,\;\; \Rightarrow \left( {1 + {k^2}} \right){x^3} + \left( {3{k^2}b - 3ak} \right){x^2} + \left( {3k{b^2} - 3ab} \right)x + {b^3} = 0.\]

Equating the coefficients of two terms of the highest degree to zero, we find the parameters of the asymptote \(k\) and \(b:\)

\[ \left\{ \begin{array}{l} 1 + {k^3} = 0\\ 3{k^2}b - 3ak = 0 \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} {k^3} = - 1\\ 3k\left( {kb - a} \right) = 0 \end{array} \right.,\;\; \Rightarrow k = - 1,\;\; \Rightarrow - b - a = 0,\;\; \Rightarrow b = - a.\]

Thus, the folium of Descartes has an oblique asymptote (Figure \(23\)), which is described by the equation \(y = -x -a.\)

Asymptote of the folium of Descartes
Figure 23.

Verify the existence of a vertical asymptote. Let its equation be written as \(y = c.\) Substitute it into the original implicit equation of the curve:

\[{x^3} + {y^3} - 3axy = 0,\;\; \Rightarrow {c^3} + {y^3} - 3acy = 0,\;\; \Rightarrow {y^3} - 3acy + {c^3} = 0.\]

Note that there is the term of the highest degree \({y^3}\) in the last equation. This means that a necessary condition for the existence of a vertical asymptote is not satisfied. Consequently, the folium of Descartes has only oblique asymptote found above.

Example 22.

Find the asymptotes of the cissoid of Diocles defined by the equation \[{y^2}\left( {2a - x} \right) = {x^3}, a \gt 0.\]

Solution.

Examine the oblique asymptotes. Substituting \(y = kx + b\) into the given equation yields:

\[{y^2}\left( {2a - x} \right) = {x^3},\;\; \Rightarrow {\left( {kx + b} \right)^2}\left( {2a - x} \right) = {x^3},\;\; \Rightarrow \left( {{k^2}{x^2} + 2kbx + {b^2}} \right)\left( {2a - x} \right) = {x^3},\;\; \Rightarrow \color{red}{2a{k^2}{x^2}} + \color{green}{4akbx} + 2a{b^2} - \color{blue}{{k^2}{x^3}} - \color{red}{2kb{x^2}} - \color{green}{{b^2}x} - \color{blue}{x^3} = 0,\;\; \color{black}{\Rightarrow} \color{blue}{\left( { - {k^2} - 1} \right){x^3}} + \color{red}{\left( {2a{k^2} - 2kb} \right){x^2}} + \color{green}{\left( {4akb - {b^2}} \right)x} + 2a{b^2} = 0.\]

The parameters \(k\) and \(b\) of the oblique asymptote are determined from the condition

\[\left\{ \begin{array}{l} - {k^2} - 1 = 0\\ 2a{k^2} - 2kb = 0 \end{array} \right..\]

However, it can be seen from the first equation that \(k\) has no real solutions:

\[- {k^2} - 1 = 0,\;\; \Rightarrow {k^2} = - 1,\;\; \Rightarrow k \in \emptyset .\]

Hence, the cissoid does not have oblique asymptotes.

Consider a vertical asymptote. Substituting \(x = c\) in the original equation, we have

\[{{y^2}\left( {2a - c} \right) = {c^3}}\;\;\text{or}\;\;\left( {2a - c} \right){y^2} - {c^3} = 0.\]

Note that the cissoid is a curve of the third order. In the latter relation, there is no term of the third order. In this case, there is a vertical asymptote and its parameter \(c\) is determined from the condition

\[2a - c = 0,\;\; \Rightarrow c = 2a.\]

Thus, the cissoid of Diocles has the vertical asymptote \(x = 2a\) (Figure \(24\)).

Asymptote of the cissoid of Diocles
Figure 24.

Note:

The equation of the cissoid can be written in explicit form:

\[{y^2}\left( {2a - x} \right) = {x^3},\;\; \Rightarrow {y^2} = \frac{{{x^3}}}{{2a - x}},\;\; \Rightarrow y = \pm \sqrt {\frac{{{x^3}}}{{2a - x}}},\]

which immediately shows that the point \(x = 2a\) has a discontinuity of the second kind, and therefore, the straight line \(x = 2a\) is a vertical asymptote.

Example 23.

Prove that the functions

\[f\left( x \right) = \sqrt {{x^2} + 4x + 3} \;\;\text{and}\;\;g\left( x \right) = \frac{{{x^3} + 2{x^2}}}{{{x^2} + 1}}\]

are asymptotically equal to each other as \(x \to +\infty.\) Estimate the error of the approximate equality \(f\left( {100} \right) \approx g\left( {100} \right).\)

Solution.

Determine the oblique asymptotes of the two functions. For the function \(f\left( x \right),\) we obtain:

\[{k_1} = \lim\limits_{x \to \infty } \frac{{f\left( x \right)}}{x} = \lim\limits_{x \to \infty } \frac{{\sqrt {{x^2} + 4x + 3} }}{x} = \lim\limits_{x \to \infty } \sqrt {\frac{{{x^2} + 4x + 3}}{{{x^2}}}} = \lim\limits_{x \to \infty } \sqrt {1 + \frac{4}{x} + \frac{3}{{{x^2}}}} = 1,\]
\[{b_1} = \lim\limits_{x \to \infty } \left[ {f\left( x \right) - {k_1}x} \right] = \lim\limits_{x \to \infty } \left( {\sqrt {{x^2} + 4x + 3} - x} \right) = \lim\limits_{x \to \infty } \frac{{\cancel{x^2} + 4x + 3 - \cancel{x^2}}}{{\sqrt {{x^2} + 4x + 3} + x}} = \lim\limits_{x \to \infty } \frac{{4 + \frac{3}{x}}}{{\sqrt {1 + \frac{4}{x} + \frac{3}{{{x^2}}}} + 1}} = 2.\]

Thus, the equation of the oblique asymptote for the first function \(f\left( x \right)\) is written as

\[y = {k_1}x + {b_1} = x + 2.\]

Consider the function \(g\left( x \right)\) and calculate the coefficients of its oblique asymptote:

\[{k_2} = \lim\limits_{x \to \infty } \frac{{g\left( x \right)}}{x} = \lim\limits_{x \to \infty } \frac{{{x^3} + 2{x^2}}}{{\left( {{x^2} + 1} \right)x}} = \lim\limits_{x \to \infty } \frac{{{x^3} + 2{x^2}}}{{{x^3} + x}} = \lim\limits_{x \to \infty } \frac{{1 + \frac{2}{x}}}{{1 + \frac{1}{{{x^2}}}}} = 1,\]
\[{b_2} = \lim\limits_{x \to \infty } \left[ {g\left( x \right) - {k_2}x} \right] = \lim\limits_{x \to \infty } \left( {\frac{{{x^3} + 2{x^2}}}{{{x^2} + 1}} - x} \right) = \lim\limits_{x \to \infty } \frac{{{x^3} + 2{x^2} - x\left( {{x^2} + 1} \right)}}{{{x^2} + 1}} = \lim\limits_{x \to \infty } \frac{{\cancel{x^3} + 2{x^2} - \cancel{x^3} - x}}{{{x^2} + 1}} = \lim\limits_{x \to \infty } \frac{{2{x^2} - x}}{{{x^2} + 1}} = \lim\limits_{x \to \infty } \frac{{2 - \frac{1}{x}}}{{1 + \frac{1}{{{x^2}}}}} = 2.\]

Consequently, the equation of the asymptote for \(g\left( x \right)\) has the same form as for \(f\left( x \right):\)

\[y = {k_2}x + {b_2} = x + 2.\]

Thus, both functions \(f\left( x \right)\) and \(g\left( x \right)\) are asymptotically equal as \(x \to +\infty.\) Calculate their values at \(x = 100:\)

\[f\left( {100} \right) = \sqrt {{{100}^2} + 4 \cdot 100 + 3} = \sqrt {10,403} \approx 101,995;\]
\[g\left( {100} \right) = \frac{{{{100}^3} + 2 \cdot {{100}^2}}}{{{{100}^2} + 1}} = \frac{{1020000}}{{10001}} \approx 101,990.\]

The relative error of the approximate equality \(f\left( {100} \right) \approx g\left( {100} \right)\) is about equal to \(5 \cdot {10^{ - 5}}.\)

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