Find the Asymptotes of the function $f(x)=3^x / (3^x+1)$
No way for Vertical asymptotes since the denominator can not be zero
Also, there is no slant asymptote since we will have horizontal asymptotes ( this is the only reason I have )
we are left with horizontal asymptote, there are two : I found one but I could not find the other
I took the limit as x approches $infinite$, sloving it using $L'H Rule$
I got $y=1$ "V.A" ,, and there is another one which is $y=0$ .
How to find it ?
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$\begingroup$There are $2$ horizontal asymptotes: $y =0$ and $y = 1$ that correspond to $x \to -\infty$, and $x \to +\infty$. To see it clearer we write: $y = 1-\dfrac{1}{3^x+1}$
$\endgroup$ 2 $\begingroup$Common sense: In the first place, recall that $3^x\to0$ as $x\to-\infty$. And if you don't recall that, look at this: \begin{align} 3^0 & = 1 \\ 3^{-1} & = 1/3 \\ 3^{-2} & = 1/9 \\ & {}\ \ \vdots \end{align} every time $x$ gets one step closer to $-\infty$, then $3^x$ gets $1/3$ as big. So it approaches $0$.
That gives us $\dfrac{3^x}{1+3^x}\to\dfrac 0 {1+0}$ as $x\to-\infty$.
Next recall that $3^x\to\infty$ as $x\to\infty$. That means the "$1$" in $\dfrac{3^x}{1+3^x}$ becomes negligible by comparison to the immense number $3^x$, so the whole thing approaches $1$. Or if you like, do this: $$ \frac{3^x}{1+3^x} = \frac{1}{\frac1{3^x}+1} \to \frac 1 {0+1} \text{ as }x\to \infty. $$
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