Find all horizontal asymptote(s) of the function $\displaystyle f(x) = \frac{x^2-x}{x^2-6x+5}$ and justify the answer by computing all necessary limits.
Also, find all vertical asymptotes and justify your answer by computing both (left/right) limits for each asymptote.
MY ANSWER so far..
For the Horizontal asymptote, I simply looked at the coefficients for both the numerator and the denominator. Both are $1$ so $\frac{1}{1}$ gives me $y=1$ as the Horizontal asymptote however I don't know how I would justify it with a limit. If i take the limit of $f(x)$, what will $x$ approach? $\infty$ or $-\infty$?
For the vertical asymtote, I set the denominator equal to $0$ and got $x=5$ and $x=1$ as the vertical asymptotes. However, I dont know how I would justify my answer using limits..
$\endgroup$ 13 Answers
$\begingroup$A better way to justify that the only horizontal asymptote is at $y = 1$ is to observe that: $$ \lim_{x \to -\infty} f(x) = \lim_{x \to \infty} f(x) = 1 $$ There is indeed a vertical asymptote at $x = 5$. To justify this, we can use either of the following two facts: \begin{align*} \lim_{x \to 5^-} f(x) &= -\infty \\ \lim_{x \to 5^+} f(x) &= \infty \\ \end{align*} Note that there is no vertical asymptote at $x = 1$; instead, the discontinuity here is just a hole. Indeed, notice that: $$ \lim_{x \to 1} \frac{x^2-x}{x^2-6x+5} = \lim_{x \to 1} \frac{x(x - 1)}{(x - 1)(x - 5)} = \lim_{x \to 1} \frac{x}{x - 5} = \frac{1}{1 - 5} = \frac{-1}{4} \neq \pm \infty $$
$\endgroup$ $\begingroup$For your horizontal asymptote divide the top and bottom of the fraction by $x^2$: $f(x) = \displaystyle \frac{1 - \frac1{x}}{1 - \frac{6}{x} + \frac{5}{x^2}}$
Then take the limit, it should be $y\to1, x\to \pm \infty$.
$\endgroup$ $\begingroup$You need to simplify your rational function to get the vertical asymptotes:
$$ f(x) = \frac{x^2 - x}{x^2 - 6x + 5} = \frac{x(x - 1)}{(x - 1)(x - 5)} \equiv \frac{x}{x - 5} $$
So right away we know that the vertical asymptote is @ $x = 5$, the horizontal asymptote is $y = 1$ and there is a removable discontinuity at $x = 1$ (that's the part that canceled).
To prove the horizontal asymptote, we just divide out the simplified part:
$$ \lim_{x \rightarrow \infty}\frac{x}{x - 5} = \lim_{x \rightarrow \infty}\frac{x\cdot1}{x\left(1 - \frac{5}{x}\right)} = \lim_{x \rightarrow \infty}\frac{1}{1 - \frac{5}{x}} = \frac{1}{1 - 0} = \frac{1}{1} = 1 $$
The same will apply for $x \rightarrow -\infty$.
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