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?
Best Answer
Yes, your solution is right.
Note that a rational function can't have both horizontal and oblique asymptotes.