Why is the vertical asymptote only here for 2?

Question

Sorry for crappy phrasing. I just need a quick refresher on an equation I took from my teacher's worksheet. Simple equation,

$$f(x) = \frac{(x-5)}{(x-5)(x-2)}$$

Why is there only an asymptote at $2$ and not $5$? Thanks in advance.

Answer 1

You're mixing up vertical asymptotes with removable discontinuities.

The function

$$f(x)=\frac{(x-5)}{(x-5)(x-2)}$$

has a removable discontinuity at $x=5$ and a vertical asymptote at $x=2$. The reason why the discontinuity is removable is because we can write

$$f(x)=\frac{(x-5)}{(x-5)(x-2)}=\frac{1}{(x-2)},x\neq5$$

by cancelling the $(x-5)$ in the numerator and denominator. The function itself isn't defined at $x=5$, although the single point of discontinuity can be removed by cancelling the common term in the numerator and denominator. The same cannot be done to the $(x-2)$, so it remains a vertical asymptote. Visually, the removable discontinuity is a hole while the vertical asymptote is a vertical line which the function cannot cross as shown by Desmos.

Answer 2

Because $\frac{x-5}{x-5}=1$ so your equation actually simplifies to $\frac{1}{x-2}$ with the only difference that $x$ cannot be $5$ because you would be dividing by zero.

If this was not the case, $f(x)=(x)$ would have asymptotes as it can be writtien like $f(x)=\frac{x(x-5)}{x-5}$

Answer 3

In $$f(x)=\frac{x-5}{(x-5)(x-2)}$$ we can divide out the $x-5$ unless $x=5$ so we have $$f(x)=\begin {cases} \text{undefined}&x=5\\\frac 1{x-2}&x \neq 5 \end {cases}$$ When you have a factor like $x-5$ (to the same power) in both the numerator and denominator it makes a removable singularity. It is a hole in the graph, but you can fill the hole by asking the function be continuous and taking the limit as you approach $5$. The zero in the numerator cancels the zero in the denominator, so the function does not go off to infinity.

Answer 4

Vertical asymptote for some function $f(x)$ in point $x=a$ occurs when $ \lim_{x\to a}f(x)=\pm\infty$

In your case $ \lim_{x\to 5}f(x)=1/3$ ,you can calculate this limit by using L'Hospitals rule for limits.

Yes when $x\to5$ denominator tends towards infinity, but so does numerator, approximately 3 times slower then denominator, that is why this limit is $1/3$.

$(x-5)$ and $(x-5)$ always have the same value no matter how x is close to a $5$, so when they divide its always equal to $1$, of course except when $x=5$, then that division doesn't make sense and hence we say that function is undefined for $x=5$ and leave it to the philosophers.