Question 674273
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Factor the numerator and denominator polynomials:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ \frac{x^2\ -\ 1}{x^2\ -\ x\ -\ 2}\ =\ \frac{(x\ -\ 1))x\ +\ 1)}{(x\ -\ 2)(x\ +\ 1)]


Values that make the denominator equal to zero:  2, -1, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \text{dom }\left{x\,\in\,\mathbb{R}\,:\,x\,\neq\,2,\,x\,\neq\,-1\right}]


Since the factor *[tex \LARGE x\ -\ 2] only occurs in the denominator, there is a vertical asymptote at


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 2]


Since the factor *[tex \LARGE x\ +\ 1] occurs in both the numerator and the denominator, there is a hole at:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(-1,\lim_{x\rightarrow-1}\,f(x)\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow-1}\,f(x)\ =\ \frac{\infty}{\infty}]


which is a L'Hôpital determinate form, so take the first derivatives of the numerator and denominator functions:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow-1}\,f(x)\ =\ \lim_{x\rightarrow-1}\,\frac{\frac{d}{dx}(x^2\,-\,1)}{\frac{d}{dx}(x^2\,-\,x\,-\,2)}\ =\  \lim_{x\rightarrow-1}\,\frac{2x}{2x\ -\ 1}\ =\ \frac{2}{3}]


So the hole is at:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(-1,\frac{2}{3}\right)]


Rational functions where the degree of the numerator polynomial is equal to the degree of the denominator polynomial have a horizontal asymptote at *[tex \LARGE  y\ =\ \frac{p}{q}] where *[tex \LARGE p] is the lead coefficient of the numerator polynomial and *[tex \LARGE q] is the lead coefficient of the denominator polynomial.  Since this rational function has a horizontal asymptote, it cannot have a slant asymptote.


Left behavior:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow-\infty}\,f(x)\ =\ \lim_{x\rightarrow-\infty}\,\frac{x^2\,-\,1}{x^2\,-\,x\,-\,2}\ =\ frac{\infty}{\infty}]


Again, we have the L'Hôpital determinate form, so take the first derivatives of the numerator and denominator functions:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow-\infty}\,f(x)\ =\ \lim_{x\rightarrow-\infty}\,\frac{\frac{d}{dx}(x^2\,-\,1)}{\frac{d}{dx}(x^2\,-\,x\,-\,2)}\ =\  \lim_{x\rightarrow-\infty}\,\frac{2x}{2x\ -\ 1}\ =\ \frac{\infty}{\infty}]


We still have the determinate form, so take the 2nd deriviative:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow-\infty}\,f(x)\ =\ \lim_{x\rightarrow-\infty}\,\frac{\frac{d^2}{dx^2}(x^2\,-\,1)}{\frac{d^2}{dx^2}(x^2\,-\,x\,-\,2)}\ =\  \lim_{x\rightarrow-\infty}\,\frac{2}{2}\ =\ 1]


Hence the function tends to the value 1 as the independent variable decreases without bound.


Similar analysis, left as an exercise for the student, is sufficient to describe the Right behavior.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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