# The idea of having a limit as $x \rightarrow a$ where the denominator is 0 confuses me

What does that mean? Say as $$x \rightarrow 2$$, our equation looks like this, $$2x + 1 \over {x - 2}$$.

Then the denominator will go towards 0. Now how would I find the limit of this? How do I know it will exist?

Also, can someone help me know why the limit of $$e^x \over {e^x + 1}$$ as $$x \rightarrow -\infty$$ is $$0$$? and when it's approaching $$+ \infty$$ the limit is 1? And why that represents the horizontal asymptote?

First off, $$\dfrac{2x+1}{x-2}$$ is not an "equation" since there is no equal sign. You probably want to call that a mathematical expression or a function (and then you "take the limit" of the function as $$x\rightarrow 2$$).

In order for the "bilateral" limit as $$x\rightarrow 2$$ to exist, the "unilateral" limits (left and right) must exist and be equal.

Replacing $$x$$ by $$2^-$$ yields $$2^--2=0^-$$ in which case your function takes the form $$\dfrac{\text{+}}{0^-}\rightarrow-\infty$$

Replacing $$x$$ by $$2^+$$ yields $$2^+-2=0^+$$ in which case your function takes the form $$\dfrac{\text{+}}{0^+}\rightarrow+\infty$$

The unilateral limits are not equal (I like to say they "don't agree") so the limit as $$x\rightarrow2$$ does not exist.

Another suggestion for the second part:

$$f(x)=\frac{e^x}{e^x+1}=\frac{e^x}{e^x(1+e^{-x})}=\frac{1}{1+e^{-x}}$$

If $$x\rightarrow+\infty$$, then $$f(x)\rightarrow\frac{1}{1+e^{-\infty}}=\frac{1}{1+0}=1$$

If $$x\rightarrow-\infty$$, then $$f(x)\rightarrow\frac{1}{1+e^{-(-\infty))}}=\frac{1}{1+\infty}=0$$

To clarify, if you have $$\frac{0}{0}$$ or $$\pm\frac{\infty}{\infty}$$, it's indeterminate and factoring might help. Otherwise:

$$\frac{+}{0^+}\rightarrow+\infty,\quad\frac{-}{0^+}\rightarrow-\infty,\quad\frac{+}{0^-}\rightarrow-\infty,\quad\frac{-}{0^-}\rightarrow+\infty,$$

• So, those are the horizontal limits correct? For the bottom example? And so when we plug in infinity into $e^x$ we just simulate what will happen to those specific functions? – ming Dec 10 '18 at 6:31
• Also can you let me know if I have this rule understood correctly? We only use the unilateral and bilateral limit rule if x is approaching a certain point to find what the limit is @ point "a"? Otherwise we don't use that rule? – ming Dec 10 '18 at 6:34
• @ming More specifically, these last 2 limits indicate the behavior of the function near $+\infty$ and $-\infty$ respectively and here. Here, the behavior is asymptotic so the function as horizontal asymptotes ($y=0$ and $y=1$). The last part where I plug-in $\infty$ might be a bit sketchy but I find it helpful to understand why the limits both equal $1$ and $0$. – orion2112 Dec 10 '18 at 6:36
• @ming You can't use the left/right limit when $x\rightarrow\pm\infty$ (because you're approaching infinity from only one possible direction, you can't approach it "from further"). What I meant by "bilateral" limit is just the limit itself (approaching the value from both left and right at the same time). This exists if the unilateral limits exist and are equal. We mainly "check" for the unilateral limits if we see a division by $0$ or for functions defined by parts. – orion2112 Dec 10 '18 at 6:45
• Cool! So only use both sided limits if we see division by 0 or a piecewise function. – ming Dec 10 '18 at 6:48

For $$\frac{2x+1}{x-2}$$, it may help to sketch a plot by yourself. When $$x$$ approaches $$2$$ from above, $$\frac{2x+1}{x-2}$$ gets arbitrarily large. When $$x$$ approaches $$2$$ from below, $$\frac{2x+1}{x-2}$$ gets arbitrarily small (negative). So, the limit does not exist.

For $$\frac{e^x}{e^x+1}$$, if $$x \to -\infty$$, then $$e^x \to 0$$, so $$\frac{e^x}{e^x+1} \to \frac{0}{0+1}$$.

When taking $$x \to \infty$$, it may help to write $$\frac{e^x}{e^x+1} = 1 - \frac{1}{e^x+1}$$.