how to find an infinity limit in a fraction

Question

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?

Answer 1

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 $$\lim_{x\to-\infty} \frac{3e^{2x}+8e^x-3}{1+e^x} = \frac{0+0-3}{1+0}=-3$$

Answer 2

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.

Answer 3

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.

Answer 4

$3\mathrm e^{2x}+8\mathrm e^x-3\sim_{+\infty}3\mathrm e^{2x}$, $\;1+\mathrm e^x\sim_{+\infty}\mathrm e^x$, 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.

Answer 5

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 $

Answer 6

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

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