# Asymptotes

## Solved Problems

Click or tap a problem to see the solution.

### Example 15

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

### Example 16

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

### Example 17

Find the asymptotes of the curve defined by the parametric equations

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

### Example 18

Find the asymptotes of the curve defined by the parametric equations

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

### Example 19

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

### Example 20

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

### Example 21

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

### Example 22

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

### 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 $$f\left( {100} \right) \approx g\left( {100} \right).$$

### 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 $$x \to \infty).$$

### 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 $$y = x - \frac{1}{3}.$$

### 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 $$19.$$

### 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 $$20.$$

### 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 $$21$$).

### 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 $$22.$$

### 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 $$y = -x -a.$$

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 $$24$$).

### 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 $$5 \cdot {10^{ - 5}}.$$