MHB 6.2.25 Evaluate Limit of x to infty

[karush]

6.2.25 DOY357

Evaluate
$\displaystyle\lim_{x\to \infty}\dfrac{e^{3x}-e^{-3x}}{e^{3x}+e^{-3x}}$

sorry I'm clueless with this e stuff:(

 

[Theia]

If you find e hard, then rewrite $$e^{3x} = t$$. Now t is approaching infinity, and no more e's in this problem.

 

[Prove It]

6.2.25 DOY357

Evaluate
$\displaystyle\lim_{x\to \infty}\dfrac{e^{3x}-e^{-3x}}{e^{3x}+e^{-3x}}$

sorry I'm clueless with this e stuff:(

$\displaystyle \begin{align*} \lim_{x \to \infty} \left( \frac{\mathrm{e}^{3\,x} - \mathrm{e}^{-3\,x}}{\mathrm{e}^{3x} + \mathrm{e}^{-3\,x}} \right) &= \lim_{x \to \infty} \left[ \left( \frac{\mathrm{e}^{3\,x} - \mathrm{e}^{-3\,x}}{\mathrm{e}^{3x} + \mathrm{e}^{-3\,x}} \right) \left( \frac{\mathrm{e}^{3\,x}}{\mathrm{e}^{3\,x}} \right) \right] \\ &= \lim_{x \to \infty} \left( \frac{\mathrm{e}^{6\,x} - 1}{\mathrm{e}^{6\,x} + 1} \right) \\ &= \lim_{x \to \infty} \left( 1 - \frac{2}{\mathrm{e}^{6\,x} + 1} \right) \\ &= 1 - 0 \\ &= 1 \end{align*}$

 

[karush]

well that was interesting a very helpfull
I have a few more coming up but will start a new post