how to find an infinity limit in a fraction I don't understand how to find the limits of this expression when $x\to\infty$ and $x\to-\infty$:
$$\left(\frac{3e^{2x}+8e^x-3}{1+e^x}\right)$$
I've searched for hours. How to compute these limits?
 A: First, let's let $u = e^{x}$.  I'm just calling the function $e^{x}$ a different name: $u$.
Now, as $x \to \infty$, we know $e^{x}$ goes to $\infty$, right?  This should be clear if you look at the graph of $e^{x}$.
So, by renaming $e^{x}$ as $u$, we can rewrite the problem as:
$\lim \limits_{x \to \infty} \dfrac{3(e^{x})^{2} + 8e^{x} - 3}{1 + e^{x}} = \lim \limits_{u \to \infty} \dfrac{3u^{2} + 8u - 3}{1 + u}$.
But
$\lim \limits_{u \to \infty} \dfrac{3u^{2} + 8u - 3}{1 + u} = \lim \limits_{u \to \infty} \dfrac{(3u^{2} + 8u - 3)\cdot \dfrac{1}{u^{2}}}{(1 + u)\cdot \dfrac{1}{u^{2}}} = \lim \limits_{u \to \infty} \dfrac{3 + \dfrac{8}{u} - \dfrac{3}{u^{2}}}{\dfrac{1}{u^{2}} + \dfrac{1}{u}}$.
Now, the numerator goes to the value $3$, while the denominator goes to the value $0$, so the fraction goes to the value $\infty$.  Hope that helps.
A: Because trying to directly evaluate the limit as $x\to\infty$ gives you the indeterminate form
$$\frac{\infty}{\infty}$$
use L'Hopital's Rule to do
\begin{align}
\lim_{x\to\infty} \frac{3e^{2x}+8e^x-3}{1+e^x}&=
\lim_{x\to\infty} \frac{\frac{d}{dx}(3e^{2x}+8e^x-3)}{\frac{d}{dx}(1+e^x)}\\
&=\lim_{x\to\infty} \frac{6e^{2x}+8e^x}{e^x}\\
&=\lim_{x\to\infty}6e^x+8 = \infty
\end{align}
The limit as $x\to-\infty$ can be evaluated directly and is
$$\lim_{x\to-\infty} \frac{3e^{2x}+8e^x-3}{1+e^x} = \frac{0+0-3}{1+0}=-3$$
A: By dividing both numerator and denominator  by $ e^{x} $ we get:
$$ \displaystyle\lim_{x\to \infty} \frac{3e^{x}+8-\frac{3}{e^{x}}}{\frac{1}{ e^{x} }+1}$$
Substituting with $\infty$ in the previous equation we get:
$$\frac {\infty +8+0}{0+1} =\infty $$
So it approach $\infty$ as $ x $ goes to $\infty $
A: To give a complete answer of what everybody is telling you, note we can divide top and bottom by $e^x$ to get
$$\lim_{x\to\infty}\frac{3e^{2x}+8e^x-3}{1+e^x}=\lim_{x\to\infty}\frac{3e^x+8-3e^{-x}}{e^{-x}+1}=\frac{\infty+8-0}{0+1}=\infty$$
Note the second to last term isn't rigorous because I can't just use infinity like a number, but that's the idea.
A: $3\mathrm e^{2x}+8\mathrm e^x-3\sim_{+\infty}3\mathrm e^{2x}$, $\;1+\mathrm e^x\sim_{+\infty}\mathrm e^x$, hence
$$\frac{3e^{2x}+8e^x-3}{1+e^x}\sim_{+\infty}\frac{3\mathrm e^{2x}}{\mathrm e^x}=3\mathrm e^x \xrightarrow[x\to+\infty]{}+\infty$$
When $x\to-\infty$, as $\mathrm e^x \to 0$, this is no indeterminate form.
A: $$\lim_{x\to\infty}\frac{3e^{2x}+8e^x-3}{1+e^x}=$$
$$\lim_{x\to\infty}\frac{-3e^{-x}+8+3e^x}{1+e^{-x}}=$$
$$\lim_{x\to\infty}\left(8+3e^x\right)=$$
$$\lim_{x\to\infty}8+\lim_{x\to\infty}\left(3e^x\right)=$$
$$\lim_{x\to\infty}8+3\lim_{x\to\infty}e^x=$$
$$\lim_{x\to\infty}8+3\exp\left(\lim_{x\to\infty}x\right)=$$
$$8+3\exp\left(\lim_{x\to\infty}x\right)=\infty$$

$$\lim_{x\to-\infty}\frac{3e^{2x}+8e^x-3}{1+e^x}=$$
$$\frac{\lim_{x\to-\infty}\left(3e^{2x}+8e^x-3\right)}{\lim_{x\to-\infty}\left(1+e^x\right)}=$$
$$\frac{\lim_{x\to-\infty}\left(3e^{2x}+8e^x-3\right)}{\lim_{x\to-\infty}1+\lim_{x\to-\infty}e^x}=$$
$$\frac{\lim_{x\to-\infty}\left(3e^{2x}+8e^x-3\right)}{1+\exp\left(\lim_{x\to-\infty}x\right)}=$$
$$\frac{\lim_{x\to-\infty}\left(3e^{2x}+8e^x-3\right)}{1}=$$
$$\lim_{x\to-\infty}\left(3e^{2x}+8e^x-3\right)=$$
$$3\left(\lim_{x\to-\infty}e^{2x}\right)+8\left(\lim_{x\to-\infty}e^x\right)+\lim_{x\to-\infty}(-3)=$$
$$3\exp\left(\lim_{x\to-\infty}2x\right)+8\exp\left(\lim_{x\to-\infty}x\right)-3=$$
$$3\exp\left(2\lim_{x\to-\infty}x\right)+8\exp\left(\lim_{x\to-\infty}x\right)-3=$$
$$3\cdot 0+8\cdot 0-3=-3$$
