Question

Explain how to find the asymptotes: vertical and horizontal to the given curve:
$y = \dfrac{{{e^{2x}} + 1}}{{{e^x} - 2}}$

Answer 1

Hint : Predicting the behavior of a curve is done using asymptotes. The basic idea is that the steep curve of a graph is denoted by asymptotes. That curve looks similar to a straight line. When either of the axes that is $x$ or $y$ will tend to $\infty$ , the curve will approach a line but will not meet it. The line that will be formed when the curve and line move together to zero is the asymptote. Also remember that this implies that the asymptote and the curve are parallel to each other.

Complete step-by-step answer :
We must first write down the equation then check when the
Let us first note down what the original equation for the curve;
$y = \dfrac{{{e^{2x}} + 1}}{{{e^x} - 2}}$
Let us see when the denominator can become $0$ ;
$\Rightarrow {e^x} - 2 = 0 \\\ \Rightarrow {e^x} = 2 \\\ \Rightarrow x = \ln 2 \;$
Moving on to the calculation of vertical asymptotes by applying the limit where $x \to \ln 2,x < \ln 2$ and $x \to \ln 2,x > \ln 2$ ;
$\Rightarrow \mathop {\lim }\limits_{_{x \to \ln 2,x < \ln 2}} y = \mathop {\lim }\limits_{_{x \to \ln 2,x < \ln 2}} \dfrac{{{e^{2x}} + 1}}{{{e^x} - 2}} \\\ \\\ \Rightarrow \mathop {\lim }\limits_{_{x \to \ln 2,x < \ln 2}} y = \mathop {\lim }\limits_{_{x \to \ln 2,x < \ln 2}} \dfrac{{{e^{2.\ln 2}} + 1}}{{{e^{\ln 2}} - 2}} \\\ \\\ \Rightarrow \mathop {\lim }\limits_{_{x \to \ln 2,x < \ln 2}} y = \mathop {\lim }\limits_{_{x \to \ln 2,x < \ln 2}} \dfrac{{4 + 1}}{{{2^ - } - 2}} \\\ \\\ \Rightarrow \mathop {\lim }\limits_{_{x \to \ln 2,x < \ln 2}} y = \dfrac{5}{{{0^ - }}} = - \infty \;$
And
$\Rightarrow \mathop {\lim }\limits_{_{x \to \ln 2,x > \ln 2}} y = \mathop {\lim }\limits_{_{x \to \ln 2,x < \ln 2}} \dfrac{{{e^{2x}} + 1}}{{{e^x} - 2}} \\\ \\\ \Rightarrow \mathop {\lim }\limits_{_{x \to \ln 2,x > \ln 2}} y = \mathop {\lim }\limits_{_{x \to \ln 2,x < \ln 2}} \dfrac{{{e^{2.\ln 2}} + 1}}{{{e^{\ln 2}} - 2}} \\\ \\\ \Rightarrow \mathop {\lim }\limits_{_{x \to \ln 2,x > \ln 2}} y = \mathop {\lim }\limits_{_{x \to \ln 2,x < \ln 2}} \dfrac{{4 + 1}}{{{2^ + } - 2}} \\\ \\\ \Rightarrow \mathop {\lim }\limits_{_{x \to \ln 2,x > \ln 2}} y = \dfrac{5}{{{0^ + }}} = + \infty \;$
So we can say that the vertical asymptote will be : $y = \ln 2$
Next we move on to the calculation of horizontal asymptotes, by applying the limit $x \to + \infty$ and $x \to - \infty$ we get;
$\Rightarrow \mathop {\lim }\limits_{_{x \to + \infty }} y = \mathop {\lim }\limits_{_{x \to + \infty }} \dfrac{{{e^{2x}} + 1}}{{{e^x} - 2}} \\\ \\\ \Rightarrow \mathop {\lim }\limits_{_{x \to + \infty }} y = \mathop {\lim }\limits_{_{x \to + \infty }} \dfrac{{{e^{2x}}}}{{{e^x}}} \\\ \\\ \Rightarrow \mathop {\lim }\limits_{x \to + \infty } y = \mathop {\lim }\limits_{_{x \to + \infty }} {e^x} \\\ \\\ \Rightarrow \mathop {\lim }\limits_{x \to + \infty } y = + \infty \;$
And
$\Rightarrow \mathop {\lim }\limits_{_{x \to - \infty }} y = \mathop {\lim }\limits_{_{x \to - \infty }} \dfrac{{{e^{2x}} + 1}}{{{e^x} - 2}} \\\ \\\ \Rightarrow \mathop {\lim }\limits_{_{x \to - \infty }} y = \mathop {\lim }\limits_{_{x \to - \infty }} \dfrac{{{0^ + } + 1}}{{{0^ + } - 2}} \\\ \\\ \Rightarrow \mathop {\lim }\limits_{x \to - \infty } y = - \dfrac{1}{2} \;$
So the horizontal asymptote will be : $y = - \dfrac{1}{2}$
So, the correct answer is “ $y = - \dfrac{1}{2}$ ”.

Note : There are many applications where asymptotes can be utilized, some of them are listed as follows:
- In Algebra for rational function and for calculus and its limits of functions, they are commonly used,
- When complex equations are considered, asymptotes can be used to make approximations for those equations.
- Asymptotes are also used for designing airplanes.
- In the games flappy bird, surprisingly but with very much significance, asymptotes are used.