Asymptotes

• An asymptote of a curve $$y = f\left( x \right)$$ that has an infinite branch is called a line such that the distance between the point $$\left( {x,f\left( x \right)} \right)$$ lying on the curve and the line approaches zero as the point moves along the branch to infinity.

Asymptotes can be vertical, oblique (slant) and horizontal. A horizontal asymptote is often considered as a special case of an oblique asymptote.

Vertical Asymptote

The straight line $$x = a$$ is a vertical asymptote of the graph of the function $$y = f\left( x \right)$$ if at least one of the following conditions is true:

${\lim\limits_{x \to a – 0} f\left( x \right) = \pm \infty ,\;\;\;}\kern-0.3pt{\lim\limits_{x \to a + 0} f\left( x \right) = \pm \infty .}$

In other words, at least one of the one-sided limits at the point $$x = a$$ must be equal to infinity.

A vertical asymptote occurs in rational functions at the points when the denominator is zero and the numerator is not equal to zero (i.e. at the points of discontinuity of the second kind). For example, the graph of the function $$y = {\large\frac{1}{x}\normalsize}$$ has the vertical asymptote $$x = 0$$ (Figure $$1\text{).}$$ In this case, both one-sided limits (from the left and from the right) tend to infinity:

${\lim\limits_{x \to 0 – 0} \frac{1}{x} = – \infty ,\;\;\;}\kern-0.3pt{\lim\limits_{x \to 0 + 0} \frac{1}{x} = + \infty .}$

A function which is continuous on the whole set of real numbers has no vertical asymptotes.

Oblique Asymptote

The straight line $$y = kx + b$$ is called an oblique (slant) asymptote of the graph of the function $$y = f\left( x \right)$$ as $$x \to +\infty$$ (Figure $$2$$) if

${\lim\limits_{x \to + \infty } \left[ {f\left( x \right) – \left( {kx + b} \right)} \right]} = {0.}$

Similarly, we introduce oblique asymptotes as $$x \to -\infty.$$

The oblique asymptotes of the graph of the function $$y = f\left( x \right)$$ may be different as $$x \to +\infty$$ and $$x \to -\infty.$$

Therefore, when finding oblique (or horizontal) asymptotes, it is a good practice to compute them separately.

The coefficients $$k$$ and $$b$$ of an oblique asymptote $$y = kx + b$$ are defined by the following theorem:

A straight line $$y = kx + b$$ is an asymptote of a function $$y = f\left( x \right)$$ as $$x \to +\infty$$ if and only if the following two limits are finite:

${\lim\limits_{x \to + \infty } \frac{{f\left( x \right)}}{x} = k\;\;\text{and}\;\;\;}\kern-0.3pt {\lim\limits_{x \to + \infty } \left[ {f\left( x \right) – kx} \right] = b.}$

Proof.

Necessity. Let the straight line $$y = kx + b$$ be an asymptote of the graph of $$y = f\left( x \right)$$ as $$x \to +\infty.$$ Then the following condition is true:

$\lim\limits_{x \to + \infty } \left[ {f\left( x \right) – \left( {kx + b} \right)} \right] = 0$

or, equivalently,

${f\left( x \right) = kx + b + \alpha \left( x \right),\;\;\;}\kern-0.3pt {\text{where}\;\;\lim\limits_{x \to + \infty } \alpha \left( x \right) = 0.}$

Dividing both sides of the equation by $$x,$$ we obtain:

${\frac{{f\left( x \right)}}{x} = \frac{{kx + b + \alpha \left( x \right)}}{x},\;\;}\Rightarrow {\frac{{f\left( x \right)}}{x} = k + \frac{b}{x} + \frac{{\alpha \left( x \right)}}{x}.}$

Consequently, in the limit as $$x \to +\infty,$$ we have

${\lim\limits_{x \to + \infty } \frac{{f\left( x \right)}}{x} } = {\lim\limits_{x \to + \infty } \left[ {k + \frac{b}{x} + \frac{{\alpha \left( x \right)}}{x}} \right]} = {k,}$

${\lim\limits_{x \to + \infty } \left[ {f\left( x \right) – kx} \right] } = {\lim\limits_{x \to + \infty } \left[ {b + \alpha \left( x \right)} \right]} ={ b.}$

Sufficiency. Suppose that there are finite limits

${\lim\limits_{x \to + \infty } \frac{{f\left( x \right)}}{x} = k\;\;\;\text{and}\;\;\;}\kern-0.3pt {\lim\limits_{x \to + \infty } \left[ {f\left( x \right) – kx} \right] = b.}$

The second limit can be written as

${\lim\limits_{x \to + \infty } \left[ {f\left( x \right) – \left( {kx + b} \right)} \right]} = {0,}$

that meets the definition of an oblique asymptote. Thus, the straight line $$y = kx + b$$ is an asymptote of the function $$y = f\left( x \right).$$

Note: Similarly we can prove the theorem for the case of $$x \to \infty.$$

Horizontal Asymptote

In particular, if $$k = 0,$$ we obtain a horizontal asymptote, which is described by the equation $$y = b.$$ The theorem on necessary and sufficient conditions for the existence of a horizontal asymptote is stated as follows:

A straight line $$y = b$$ is an asymptote of a function $$y = f\left( x \right)$$ as $$x \to +\infty,$$ if and only if the following limit is finite:

$\lim\limits_{x \to + \infty } f\left( x \right) = b.$

The case $$x \to -\infty$$ is considered in the same way.

Asymptotes of a Parametrically Defined Curve

Let a plane curve be defined by the parametric equations

${x = \varphi \left( t \right),\;\;\;}\kern-0.3pt{y = \psi \left( t \right).}$

This curve has a vertical asymptote $$x = a$$ as $$t \to {t_1}$$ if the following conditions are true:

${\lim\limits_{t \to {t_1}} \varphi \left( t \right) = a\;\;\;\text{and}\;\;\;}\kern-0.3pt {\lim\limits_{t \to {t_1}} \psi \left( t \right) = \pm \infty .}$

Similarly, a parametrically defined curve has a horizontal asymptote $$y = b$$ as $$t \to {t_2}$$ if

${\lim\limits_{t \to {t_2}} \varphi \left( t \right) = \pm \infty\;\;\;\text{and}\;\;\;}\kern-0.3pt {\lim\limits_{t \to {t_2}} \psi \left( t \right) = b.}$

Here $$a$$ and $$b$$ are finite quantities.

A parametrically defined curve has an oblique (slant) asymptote $$y = kx + b$$ as $$t \to {t_3}$$ if, at this value of $$t,$$ both limits are infinite:

${\lim\limits_{t \to {t_3}} \varphi \left( t \right) = \pm \infty ,\;\;\;}\kern-0.3pt {\lim\limits_{t \to {t_3}} \psi \left( t \right) = \pm \infty ,}$

and the coefficients $$k$$ and $$b$$ have finite values:

${k = \lim\limits_{t \to {t_3}} \frac{{\psi \left( t \right)}}{{\varphi \left( t \right)}},\;\;\;}\kern-0.3pt {b = \lim\limits_{t \to {t_3}} \left[ {\psi \left( t \right) – k\varphi \left( t \right)} \right].}$

Asymptote of a Polar Curve

Consider a curve given in polar coordinates by the equation

${\rho = \rho \left( \varphi \right)\;\;\;\text{or}\;\;\;}\kern-0.3pt{F\left( {\rho ,\varphi } \right) = 0.}$

Its asymptote (if it exists) can be described by the two parameters: the distance $$p$$ from the center to the asymptote (segment $$OA$$ in Figure $$3\text{)}$$ and the angle $$\alpha$$ of inclination of the asymptote to the polar axis.

These parameters $$\alpha$$ and $$p$$ are defined by the formulas

${\alpha = \lim\limits_{\rho \to \infty } \varphi ,\;\;\;}\kern-0.3pt {p = \lim\limits_{\rho \to \infty } \left[ {\rho \sin \left( {\alpha – \varphi } \right)} \right].}$

Notes:

1. In the last formula, the limit transition $$\rho \to \infty$$ can be replaced by an equivalent condition $$\varphi \to \alpha;$$
2. The parameter $$p$$ can be both positive and negative.

Asymptotes of an Implicit Curve

An implicitly defined algebraic curve is described by the equation

$F\left( {x,y} \right) = 0,$

where the left-hand side is a polynomial in the variables $$x$$ and $$y.$$

In differential geometry, the following method for finding an oblique asymptote of an algebraic curve is used. Suppose that the asymptote is described by the equation $$y = kx + b.$$ Substituting this expression for $$y$$ in the equation of the curve, we obtain an algebraic equation in one variable $$x:$$

${{A_0}{x^n} + {A_1}{x^{n – 1}} + \ldots} + {{A_{n – 1}}x + {A_n} = 0,}$

where the coefficients $${A_i}$$ depend on the parameters of the asymptote $$k$$ and $$b$$ (the coefficient $${A_0}$$ depends only on $$k$$). The values of $$k$$ and $$b$$ can be determined from the condition:

$\left\{ \begin{array}{l} {A_0}\left( k \right) = 0\\ {A_1}\left( {k,b} \right) = 0 \end{array} \right..$

To find a vertical asymptote, it is necessary to substitute its equation $$x = a$$ in the equation of the curve and convert the latter to the form:

${{B_0}{y^n} + {B_1}{y^{n – 1}} + \ldots} + {{B_{n – 1}}y + {B_n} = 0.}$

A necessary condition for the existence of a vertical asymptote is the absence of the term of the highest degree $${B_0}{y^n}$$ in the last equation.

The value of the parameter $$a$$ is determined from the condition

${B_1}\left( a \right) = 0.$

The above formulas for the asymptotes of an implicit curve are valid if the curve has no singular points at infinity.

• Solved Problems

Example 1

Find the asymptotes of the function $f\left( x \right) = \frac{x}{{x + 1}}.$

Example 2

Find the asymptotes of the function $f\left( x \right) = {\frac{{2x – 3}}{{x + 1}}}.$

Example 3

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

Example 4

Find the asymptotes of the function $f\left( x \right) = {\frac{{{x^2} – 2x – 3}}{x}}.$

Example 5

Find the asymptotes of the function $f\left( x \right) = \frac{{3{x^2} – 2x + 1}}{{x – 1}}.$

Example 6

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

Example 7

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

Example 8

Find the asymptotes of the function $f\left( x \right) = {\frac{{{x^3}}}{{{x^2} + 1}}}.$

Example 9

Find the asymptotes of the function $f\left( x \right) = {\frac{{{x^2} – 3x + 2}}{{{x^2} + 3x + 2}}}.$

Example 10

Find the asymptotes of the function $f\left( x \right) = x + \arctan x.$

Example 11

Find the asymptotes of the hyperbolic tangent $f\left( x \right) = \tanh x.$

Example 12

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

Example 13

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

Example 14

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

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 ${x = \varphi \left( t \right) = \frac{1}{{t – 2}},\;\;\;}\kern-0.3pt {y = \psi \left( t \right) = \frac{t}{{t – 1}}.}$

Example 18

Find the asymptotes of the curve defined by the parametric equations ${x = \varphi \left( t \right) = \frac{t}{{1 – {t^2}}},\;\;}\kern-0.3pt {y = \psi \left( t \right) = \frac{{{t^2}}}{{1 – {t^2}}}}$

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 ${{y^2}\left( {2a – x} \right) = {x^3},\;\;\;}\kern-0.3pt{a \gt 0.}$

Example 23

Prove that the functions ${{f\left( x \right) = \sqrt {{x^2} + 4x + 3} \;\;\;\text{and}}\;\;\;}\kern-0.3pt {g\left( x \right) = \frac{{{x^3} + 2{x^2}}}{{{x^2} + 1}}}$ 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 1.

Find the asymptotes of the function $f\left( x \right) = \frac{x}{{x + 1}}.$

Solution.

When $$x = -1,$$ the function has a discontinuity of the second kind. Indeed:

${\lim\limits_{x \to – 1 – 0} f\left( x \right) } = {\lim\limits_{x \to – 1 – 0} \frac{x}{{x + 1}} } = {\frac{{ – 1}}{{\left( { – 1 – 0} \right) + 1}} } = {\frac{{ – 1}}{{ – 0}} = + \infty ,}$

${\lim\limits_{x \to – 1 + 0} f\left( x \right) } = {\lim\limits_{x \to – 1 + 0} \frac{x}{{x + 1}} } = {\frac{{ – 1}}{{\left( { – 1 + 0} \right) + 1}} } = {\frac{{ – 1}}{{ + 0}} = – \infty .}$

Hence, $$x = -1$$ is the equation of the vertical asymptote.

Find the horizontal asymptote. Compute the limit:

${\lim\limits_{x \to \pm \infty } \frac{x}{{x + 1}}} ={ \lim\limits_{x \to \pm \infty } \frac{1}{{1 + \frac{1}{x}}} }={ 1.}$

Thus, there exists a horizontal asymptote for the curve, and its equation is $$y = 1.$$

The function has no oblique asymptotes. This can be verified by calculating the coefficients $$k$$ and $$b:$$

${k = \lim\limits_{x \to \pm \infty } \frac{{y\left( x \right)}}{x} } = {\lim\limits_{x \to \pm \infty } \frac{x}{{\left( {x + 1} \right)x}} } = {\lim\limits_{x \to \pm \infty } \frac{x}{{{x^2} + x}} } = {\lim\limits_{x \to \pm \infty } \frac{{\frac{1}{x}}}{{1 + \frac{1}{x}}} = 0,}$

${b = \lim\limits_{x \to \pm \infty } \left[ {y\left( x \right) – kx} \right] } = {\lim\limits_{x \to \pm \infty } \left( {\frac{x}{{x + 1}} – 0} \right) } = {\lim\limits_{x \to \pm \infty } \frac{1}{{1 + \frac{1}{x}}} = 1.}$

It can be seen that actually we obtained the horizontal asymptote, which has already been defined above.

So, the graph of the function has the vertical asymptote $$x = -1$$ and the horizontal asymptote $$y = 1$$ (Figure $$4$$).

Example 2.

Find the asymptotes of the function $f\left( x \right) = {\frac{{2x – 3}}{{x + 1}}}.$

Solution.

The function is defined for all $$x \ne -1.$$ Compute the one-sided limits at $$x = -1:$$

${\mathop {\lim }\limits_{x \to – 1 – 0} f\left( x \right) }={ \mathop {\lim }\limits_{x \to – 1 – 0} \frac{{2x – 3}}{{x + 1}} }={ \frac{{ – 5}}{{ – 0}} }={ + \infty ;}$

${\mathop {\lim }\limits_{x \to – 1 + 0} f\left( x \right) }={ \mathop {\lim }\limits_{x \to – 1 + 0} \frac{{2x – 3}}{{x + 1}} }={ \frac{{ – 5}}{{ + 0}} }={ – \infty .}$

Hence, $$x = -1$$ is a vertical asymptote of the function.

Find the possible horizontal asymptotes at $$x \to \pm\infty:$$

${\mathop {\lim }\limits_{x \to \pm \infty } f\left( x \right) }={ \mathop {\lim }\limits_{x \to \pm \infty } \frac{{2x – 3}}{{x + 1}} }={ \mathop {\lim }\limits_{x \to \pm \infty } \frac{{2 – \frac{3}{x}}}{{1 + \frac{1}{x}}} }={ 2.}$

We obtained a horizontal asymptote at $$y = 2.$$

Check for oblique (slant) asymptotes by calculating the slope $$k:$$

${k = \mathop {\lim }\limits_{x \to + \infty } \frac{{f\left( x \right)}}{x} }={ \mathop {\lim }\limits_{x \to + \infty } \frac{{2x – 3}}{{x\left( {x + 1} \right)}} }={ \mathop {\lim }\limits_{x \to + \infty } \frac{{2x – 3}}{{{x^2} + x}} }={ \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{2}{x} – \frac{3}{{{x^2}}}}}{{1 + \frac{1}{x}}} }={ \frac{0}{1} }={ 0.}$

The same is for $$x \to -\infty.$$ Therefore, there is no oblique asymptotes.

The graph of the function is sketched in Figure $$5$$.

Example 3.

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

Solution.

We compute the one-sided limits at $$x = 1:$$

${\lim\limits_{x \to 1 – 0} f\left( x \right) } = {\lim\limits_{x \to 1 – 0} \frac{x}{{{{\left( {x – 1} \right)}^2}}} } = {\frac{{1 – 0}}{{{{\left( {1 – 0 – 1} \right)}^2}}} } = {\frac{1}{{{{\left( { – 0} \right)}^2}}} = + \infty ,}$

${\lim\limits_{x \to 1 + 0} f\left( x \right) } = {\lim\limits_{x \to 1 + 0} \frac{x}{{{{\left( {x – 1} \right)}^2}}} } = {\frac{{1 + 0}}{{{{\left( {1 + 0 – 1} \right)}^2}}} } = {\frac{1}{{{{\left( { + 0} \right)}^2}}} = + \infty.}$

The limits are equal to infinity. Therefore, $$x = 1$$ is a vertical asymptote of the curve.

Next, let us examine the oblique asymptotes:

$\require{cancel} {k = \lim\limits_{x \to \pm \infty } \frac{{f\left( x \right)}}{x} } = {\lim\limits_{x \to \pm \infty } \frac{\cancel{x}}{{{{\left( {x – 1} \right)}^2}\cancel{x}}} } = {\lim\limits_{x \to \pm \infty } \frac{1}{{{{\left( {x – 1} \right)}^2}}} = 0;}$

${b = \lim\limits_{x \to \pm \infty } \left[ {f\left( x \right) – kx} \right] } = {\lim\limits_{x \to \pm \infty } \left[ {\frac{x}{{{{\left( {x – 1} \right)}^2}}} – 0} \right] } = {\lim\limits_{x \to \pm \infty } \frac{x}{{{x^2} – 2x + 1}} } = {\lim\limits_{x \to \pm \infty } \frac{{\frac{1}{x}}}{{1 – \frac{2}{x} + \frac{1}{{{x^2}}}}} = 0.}$

This shows that instead of an oblique asymptote there is a horizontal asymptote as $$x \to \pm \infty.$$ Its equation is $$y = 0,$$ i.e. the asymptote is the $$x$$-axis.

A view of the function and its asymptotes is given in Figure $$6.$$

Example 4.

Find the asymptotes of the function $f\left( x \right) = {\frac{{{x^2} – 2x – 3}}{x}}.$

Solution.

The function has a discontinuity at $$x = 0.$$ Since

${\mathop {\lim }\limits_{x \to 0 – 0} f\left( x \right) }={ \mathop {\lim }\limits_{x \to 0 – 0} \frac{{{x^2} – 2x – 3}}{x} }={ + \infty ;}$

${\mathop {\lim }\limits_{x \to 0 + 0} f\left( x \right) }={ \mathop {\lim }\limits_{x \to 0 + 0} \frac{{{x^2} – 2x – 3}}{x} }={ – \infty ,}$

the straight line $$x = 0$$ (the $$y-$$axis) is a vertical asymptote.

The function does not have a horizontal asymptote because the following limits are infinite:

${\mathop {\lim }\limits_{x \to + \infty } f\left( x \right) }={ \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^2} – 2x – 3}}{x} }={ + \infty ;}$

${\mathop {\lim }\limits_{x \to – \infty } f\left( x \right) }={ \mathop {\lim }\limits_{x \to – \infty } \frac{{{x^2} – 2x – 3}}{x} }={ – \infty .}$

We write the function as follows:

${f\left( x \right) }={ \frac{{{x^2} – 2x – 3}}{x} }={ x – 2 – \frac{3}{x}.}$

The term $$\large{\frac{3}{x}}\normalsize$$ approaches zero as $$x \to \pm \infty.$$ Hence, the function has the oblique asymptote $$y = x – 2.$$

Example 5.

Find the asymptotes of the function $f\left( x \right) = \frac{{3{x^2} – 2x + 1}}{{x – 1}}.$

Solution.

It is clear that the line $$x = 1$$ is a vertical asymptote, because at this point the function has a discontinuity and the following conditions are true:

${\lim\limits_{x \to 1 – 0} f\left( x \right) } = {\lim\limits_{x \to 1 – 0} \frac{{3{x^2} – 2x + 1}}{{x – 1}} } = {\frac{{3{{\left( {1 – 0} \right)}^2} – 2\left( {1 – 0} \right) + 1}}{{1 – 0 – 1}} }= -{ \infty ,}$

${\lim\limits_{x \to 1 + 0} f\left( x \right) } = {\lim\limits_{x \to 1 + 0} \frac{{3{x^2} – 2x + 1}}{{x – 1}} } = {\frac{{3{{\left( {1 + 0} \right)}^2} – 2\left( {1 + 0} \right) + 1}}{{1 + 0 – 1}} }= +{ \infty .}$

We write the function as

${y=f\left( x \right) = \frac{{3{x^2} – 2x + 1}}{{x – 1}} } = {\frac{{3{x^2} – 3x + x – 1 + 2}}{{x – 1}} } = {\frac{{3x\cancel{\left( {x – 1} \right)}}}{\cancel{x – 1}} + \frac{\cancel{x – 1}}{\cancel{x – 1}} + \frac{2}{{x – 1}} } = {3x + 1 + \frac{2}{{x – 1}} }={ 3x + 1 + \alpha \left( x \right),}$

where $$\alpha \left( x \right) \to 0$$ as $$x \to \pm \infty.$$

Thus, the function has the oblique asymptote $$y = 3x + 1.$$

Note that a rational function may have an oblique asymptote if the degree of the numerator is one greater than the degree of the denominator. A schematic view of this curve is shown in Figure $$8.$$

Example 6.

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

Solution.

The function is defined for all $$x \in \mathbb{R}$$ except $$x = 1$$ where the denominator is equal zero. Look for vertical asymptotes:

${\mathop {\lim }\limits_{x \to 1 – 0} f\left( x \right) }={ \mathop {\lim }\limits_{x \to 1 – 0} \frac{{{x^2}}}{{x – 1}} }={- \infty ;}$

${\mathop {\lim }\limits_{x \to 1 + 0} f\left( x \right) }={ \mathop {\lim }\limits_{x \to 1 + 0} \frac{{{x^2}}}{{x – 1}} }={+ \infty .}$

Hence there is a vertical asymptote given by the equation $$x = 1.$$

Notice that the degree of the numerator is one more than the degree of the denominator. Therefore, the function has an oblique asymptote. We can find its equation by long division:

$\require{cancel}{\frac{{{x^2}}}{{x – 1}} }={ \frac{{{x^2} – x + x – 1 + 1}}{{x – 1}} }={ \frac{{x\left( {x – 1} \right) + \left( {x – 1} \right) + 1}}{{x – 1}} }={ \frac{{x\cancel{\left( {x – 1} \right)}}}{\cancel{x – 1}} + \frac{\cancel{x – 1}}{\cancel{x – 1}} + \frac{1}{{x – 1}} }={ x + 1 + \frac{1}{{x – 1}}.}$

As $$\mathop {\lim }\limits_{x \to \pm \infty } \large{\frac{1}{{x – 1}}}\normalsize = 0,$$ the equation of the oblique asymptote is written in the form

$y = x + 1.$

This oblique asymptote exists for $$x \to \pm\infty.$$ Hence, the function does not have horizontal asymptotes.

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Problems 1-6
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Problems 7-23