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1) Removable discontinuity at $x=8$; vertical asymptote at $x=7$.

2) $\lim\limits _{ x\rightarrow 0 }{ f\left( x \right) } =\frac { 2 }{ 9 }$ ; vertical asymptote at $x= -9/4$

3) Hole at $(-4, 2)$

4) Vertical asymptote at $x= 2$; hole at $(6,3)$

Could you please provide a solution to each condition?

Thanks very much

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    I am guessing those are supposed to mean removable discontinuity. I don't know how to make an equation for each requirement though.2012-07-02

3 Answers 3

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I will try to give you a general formula you can use for each of those.

The so called homographic functions, with $ad-cb \neq 0$

$y=\frac{ax+b}{cx+d}$

have the following general properties:

$f(0)=\frac{b}{d}$

$\lim\limits_{x \to -\frac d c} f(x) = \infty $ $\lim\limits_{x \to \infty} f(x) = \frac a c$

Maybe, a more natural form is the canonical one, where we have

$y=k+\frac{a}{x-h}$

The properties are now even more evident,

$\lim\limits_{x \to \infty} f(x) = k$

$\lim\limits_{x \to h} f(x) = \infty $

and $f(0)=k-\frac a h $

Their complex analog, the Möbius transformations, play an important role in complex analysis.

Also, given any continuous function $g(x)$, we can create a removable discontinuity by defining

$G(x) = \begin{cases} g(x) \text{ ; } x \neq a \cr M \text{ ; } x =a \end{cases}$

where $M \neq g(a)$.

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$y={\sin x\over x}$ has a removable discontinuity at $x=0$. Can you figure out how to move it to $x=8$? and then how to tack on a vertical aymptote at $x=7$?

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HINT

  1. My favorite vertical asymptote comes from $1/x^2$. Do you know how to move the asymptote around?
  2. My favorite removable discontinuity is the function $f(x)=0$ for all $x\neq 0$, $f(0)=1$.