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Let's suppose I have the graph y = $\frac 1x$ and that I do not know how it looks visually/graphically. I know there is an asymptote at $x = 0$, but do not know if the graph is approaching positive or negative infinity at both sides of the asymptote. I would like to know if it is possible to tell algebraically just from a given equation where the graph goes on both sides of the asymptote. I would like to do this without plugging in two values for $x$ (smaller than $x$ /greater than $x$), unless that that is the only solution to this problem.

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  • $\begingroup$ positive/postive=postive and postive/negative=negative. Division by $0$ but not intermediate form suggests the limit goes to $\infty$, $-\infty$ or $DNE$. $\endgroup$ – Ahmed S. Attaalla Jan 5 '17 at 0:38
  • $\begingroup$ If you know there is a vertical asymptote at $x=0$, the question is whether $y$ is positive or negative when $x$ is close to $0$ (on one side or the other). Depending on the specific function, it may be easy to determine this. For example, if $f(0) = 0$ and $f'(0) > 0$, then $1/f(x) > 0$ for $x$ positive and small, $<0$ for $x$ negative and small. The reverse if $f'(0) < 0$. If $f'(0) = 0$, you have to look at the least integer $j$ (if any) for which $f^{(j)}(0) \ne 0$... $\endgroup$ – Robert Israel Jan 5 '17 at 0:49
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So you know the limit on one side $\to \infty$ or $\to -\infty$ and would like to figure which one the limit tends to from the two.

This is easy just note $\frac{\text{positive}}{\text{negative}}=\text{negative}$, etc. All those basic rules come in handy.

It all depends on the function though.

Say we want to find $\lim_{\to 0^+}$

$$\frac{1}{x}$$

I think you'll agree the limit is infinite. We want to see what happens as $x>0$ but $x$ gets close to zero. Well because $x$ is positive in that case.

$$\frac{1}{x}$$

Must be also positive.

So it all just had to do with basic rules and often algebraic manipulation.

More examples,

$$\frac{1}{x^2}$$

Find the limit as $x \to 0$

$x^2>0$ so $\frac{1}{x^2}$ is positive. It diverges, so it goes to $+\infty$.

$$-\frac{1}{x^3}$$

$$\to 0^-$$

We approaching from $x<0 \implies x^3<0 \implies -\frac{1}{x^3}>0$. The limit diverges, so it must go to $+\infty$.

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  • $\begingroup$ Thanks this is what I was looking for. $\endgroup$ – user402894 Jan 6 '17 at 11:27
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Well..

for $y = x^{-c}$ where $c$ is a constant:

if $c$ is an odd integer, left side of asymptote approaches negative infinityand right side of asymptote approaches positive infinity

if $c$ is an even integer, both sides approach positive infinity

if $c$ is not an integer, then the right side of asymptote approaches positive infinity -- there is no 'left side' as that would be in the realm of imaginary numbers

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  • $\begingroup$ If $c$ is a rational with odd denominator (in lowest terms), $x^{-c}$ does have a real value for negative $x$. $\endgroup$ – Robert Israel Jan 5 '17 at 0:44
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To get a good idea of where the graph goes, look at points around where you want to study! The idea that you can get arbitrarily close to something without actually being there is called a limit.

For example, if we APPROACH x=0 from the RIGHT, f(x)=1/x

f(1) = 1 f(0.1)=10 f(0.01)=100, etc. so it is obvious that it approaches positive infinity from the right.

whereas on the left

f(-1) = -1 f(-0.1) = -10 f(-0.01) = -100, etc. so it approaches negative infinity from the left.

So you can't really say much about 1/x, because it approaches different infinities on both sides

The graph is negative on the left, and positive on the right.

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Working out "If I keep getting closer to this point on the graph does it go to infinity?" in math is called "Taking the limit". There isn't a single fixed procedure that works for every equation but a family of tools that can be used to take an equation and find out its properties.

Basically to show a graph has a limit at infinity at some point on the graph (which we will call $x_0$) you carry out a proof showing that at the no matter how high a 'y' value someone picks you can find a x value where all points closer to $x_0$ are greater then the given y value.

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