Given the curve $y=\frac{5x}{x-3}$. Find its asymptotes, if any. Given the curve $y=\frac{5x}{x-3}$. To examine its asymptotes if any.
We are only taking rectilinear asymptotes in our consideration
My solution goes like this:

We know that, a straight line $x=a$, parallel to $y$ axis can be a vertical asymptote of a branch of the curve $y=f(x)$, iff $f(x)\to\infty$ when $x\to a+0$, $x\to a-0$ or $x\to a$. Similarly, a straight line $y=a$, parallel to $x$ axis can be a horizontal asymptote of a branch of the curve $x=\phi(y)$, iff $x\to\infty$ when $y\to a+0$, $y\to a-0$ or $y\to a$. Using these lemmas, we obtain $x=3$ and $y=5$ as the rectilinear asymptotes. This is because, $y=\frac{5x}{x-3}=f(x)$, then as $x\to 3$, $f(x)\to \infty$. Also, since $y=\frac{5x}{x-3}$, thus,$xy-3y=5x$ or $xy-5x=3y$, hence $x=\frac{3y}{y-5}=\phi(y)$. Due to which $\phi(y)\to \infty$ as $y\to 5$.

Now, I tried checking, if there is any oblique asymptote. We know that, $y=mx+c$, is an oblique asymptote of $y=f(x)$, iff $\exists $ a finite $m=\lim_{|x|\to\infty}\frac{y}{x}$ and $c=\lim_{|x|\to\infty} y-mx$. Now, we have the function $y=\frac{5x}{x-3}$ and hence, $\frac{y}{x}=\frac{5}{x-3}$ and hence, $m=\lim_{|x|\to\infty}\frac{y}{x}=\lim_{|x|\to\infty}\frac{5}{x-3}=0$. Hence, $c=\lim_{|x|\to\infty} y-mx=\frac{5x}{x-3}=\frac{5}{1-\frac 3x}=5$. So, the oblique asymptote of the given curve is $y=mx+c=5$, which is same as the horizontal asymptote, parallel to $x-axis$ as shown above. Is the solution correct? If not, where is it going wrong?
 A: You have correctly found the asymptotes of the function.  Observe that we can express the function in the form
$$f(x) = \frac{5x}{x - 3} = \frac{5x - 15 + 12}{x - 3} = \frac{5x - 15}{x - 3} + \frac{12}{x - 3} = \frac{12}{x - 3} + 5$$
which tells us that the graph of $f$ can be obtained by dilating the graph of $y = 1/x$ by a factor of $12$, shifting it to the right by three units, and up by $5$ units. Since the curve $y = 1/x$ has horizontal asymptote $y = 0$ and vertical asymptote $x = 0$, it follows that the function $f$ has horizontal asymptote $y = 5$ and horizontal asymptote $x = 3$.
A function $f$ has horizontal asymptote $y = k$ if
$$\lim_{x \to \infty} f(x) = k$$
or
$$\lim_{x \to -\infty} f(x) = k$$
Observe that
\begin{align*} 
\lim_{x \to \infty} f(x) & = \lim_{x \to \infty} \frac{4x}{x - 3} = \lim_{x \to \infty} \left(\frac{12}{x - 3} + 5\right) = 0 + 5 = 5\\
\lim_{x \to -\infty} f(x) & \lim_{x \to -\infty} \frac{4x}{x - 3} = = \lim_{x \to \infty} \left(\frac{12}{x - 3} + 5\right) = 0 + 5 = 5
\end{align*}
which tells us that the only horizontal asymptote is $y = 5$.
In fact, a rational function can have at most one horizontal asymptote.  This is not true for all functions.  For instance, the function $g: \mathbb{R} \to \mathbb{R}$ defined by $g(x) = \frac{x}{|x| + 1}$ has two horizontal asymptotes since
\begin{align*}
\lim_{x \to \infty} g(x) & = \lim_{x \to \infty} \frac{x}{|x| + 1} = \lim_{x \to \infty} \frac{x}{x + 1} = \lim_{x \to \infty} \frac{1}{1 + \frac{1}{x}} = 1\\
\lim_{x \to \infty} g(x) & = \lim_{x \to \infty} \frac{x}{|x| + 1} = \lim_{x \to \infty} \frac{x}{-x + 1} = \lim_{x \to \infty} \frac{1}{-1 + \frac{1}{x}} = -1
\end{align*}
A function $f$ has a vertical asymptote at $x = a$ if at least one of the following four statements is true:
\begin{align*}
\lim_{x \to a^+} f(x) & = \infty & \lim_{x \to a^+} f(x) & = -\infty\\
\lim_{x \to a^-} f(x) & = \infty & \lim_{x \to a^-} f(x) & = -\infty
\end{align*}
In this case,
\begin{align*}
\lim_{x \to 3^+} f(x) & = \lim_{x \to 3^+} \frac{5x}{x - 3} = \lim_{x \to 3^+}  \left(\frac{12}{x - 3} + 5\right) = \infty\\
\lim_{x \to 3^-} f(x) & = \lim_{x \to 3^-} \frac{5x}{x - 3} = \lim_{x \to 3^-}  \left(\frac{12}{x - 3} + 5\right) = -\infty
\end{align*}
so $x = 3$ is a vertical asymptote of the graph.
A rational function can have a horizontal asymptote when the degree of the numerator is at most that of the denominator, which is the case here since both the numerator and denominator have degree one.  A rational function can have an oblique asymptote when the degree of the numerator exceeds that of the denominator by one.  A rational function can have a curvilinear asymptote when the degree of the numerator exceeds that of the denominator by more than one.
A: Yes, your solution is right.

*

*Horizontal asymptotes occur when the numerator of a rational function has degree less than or equal to the degree of the denominator.

*Oblique asymptotes occur when the degree of the denominator of a rational function is one less than the degree of the numerator.

Note that a rational function can't have both horizontal and oblique asymptotes.
A: A homographic function is the quotient of two first-degree polynomial functions, through an expression in the form, with $c\ne 0$,
$$f \left( x \right)=\dfrac {ax+b} {cx+d}=\frac{5x+0}{x-3}$$
$a=5, b=0, c=1, d=-3$.

*

*If $c=0$ then $y =\frac{a \cdot x}{d}+\frac{b}{d}$, which is the equation of a line of angular coefficient $a \over d$, which intersects the $y$-axis at the point of ordinate $b \over d$.

If the mixed product between the coefficients $a\cdot d=b\cdot c$, then $d=\frac{b\cdot c}{a}$ can be substituted, and then, picking up by common factor, $y=\frac{a(ax+b)}{c(ax+b)}$, which simplified gives $y=\frac{a}{c}$, i.e., a line parallel to the $x$-axis representing the horizontal asymptote of the homographic function (The same result is reached by exploiting the definition of limit, i.e., $$y=\lim_{x \to +\infty} \frac{(ax+b)}{(cx+d)} =\lim_{x \to +\infty} \frac{x(a+\frac{b}{x})}{x(c+\frac{d}{x})} = \frac{a+0}{c+0} =\frac{a}{c}$$ which is the horizontal asymptote).

*

*If $c \neq 0$ and $a \cdot d \ne  b \cdot c$, then the homographic function represents an equilateral hyperbola with asymptotes parallel to the coordinate axes. In particular, the asymptotes have equations $y=\frac{a}{c}$ and $x=-\frac{d}{c}$. The $x=-\frac{d}{c}$ is a vertical asymptote because must be $cx+d\ne 0$. In our case therefore long calculations are not there. In fact $ad-bc=5\cdot(-3)-0\ne 0$. Therefore the asymptotes have equation  $y=5$ (horizontal) and $x=-\frac{-3}{1}=3$ (vertical).

