# Thread: How to calculate this limit?

1. ## How to calculate this limit?

I cant seem to figure out what to do to solve this problem. Can anyone please help?

lim x--> infinity ((2e^-x - 6e^x) / 2e^x + e^-x)

The answer is -3 but how?

also....how would I do this one? (very similar prob)

lim x-->0+ ((2 - e^-x) / e^1/x )

2. Is this $\displaystyle \lim_{x \to \infty}\left(\frac{2e^{-x} - 6e^x}{2e^x} + e^{-x}\right)$ or $\displaystyle \lim_{x \to \infty}\frac{2e^{-x} - 6e^x}{2e^x + e^{-x}}$?

3. sorry for being unclear, it is the latter though

4. $\displaystyle \frac{2e^{-x} - 6e^x}{2e^x + e^{-x}} = \frac{\frac{2}{e^x} - 6e^x}{2e^x + \frac{1}{e^x}}$

$\displaystyle = \frac{\frac{2 - 6e^{2x}}{e^x}}{\frac{2e^{2x} + 1}{e^x}}$

$\displaystyle = \frac{2-6e^{2x}}{2e^{2x} + 1}$.

Therefore $\displaystyle \lim_{x \to \infty}\frac{2e^{-x} - 6e^x}{2e^x + e^{-x}} = \lim_{x \to \infty}\frac{2 - 6e^{2x}}{2e^{2x} + 1}$.

Since this goes to $\displaystyle \frac{\infty}{\infty}$, you can use L'Hospital's Rule...

5. Hello, colerelm1!

$\displaystyle \lim_{x\to\infty} \frac{2e^{-x} - 6e^x}{2e^x + e^{-x}}$

The function is: . $\displaystyle \frac{2e^{-x} - 6e^x}{2e^x + e^{-x}}$

Divide numerator and denominator by $e^x\!:\;\;\dfrac{2e^{-2x} - 6}{2 + e^{-2x}}$

$\displaystyle\text{Therefore: }\;\lim_{x\to\infty}\left(\frac{\frac{2}{e^{2x}} - 6}{2 + \frac{1}{e^{2x}}}\right) \;=\; \frac{0-6}{2+0} \;=\;-3$