# How do you evaluate the limit: lim_(x->oo)(e^(-2x)-e^(2x))/(e^(-2x)+e^(2x))?

Sep 15, 2017

$- 1$.

#### Explanation:

First of all, divide everything in the limit by the highest power, so in this case, ${e}^{2 x}$.
${\lim}_{\text{x->∞}} \frac{{e}^{- 4 x} - 1}{{e}^{- 4 x} + 1}$
Use the following property:
lim_"x->a"[f(x)/g(x)]=(lim_"x->a"f(x))/(lim_"x->a"g(x)) = (lim_"x->∞"e^(-4x)-1)/(lim_"x->∞"(e^(-4x)+1) = (lim_"x->∞"1/(e^(4x))-1)/(lim_"x->∞"(1/(e^(4x))+1)

Now, to solve the limit. Just by glancing at it, ${\lim}_{\text{x->∞}} \frac{1}{{e}^{4 x}}$ looks like it is basically $0$, so we can write that in.

$\frac{0 - 1}{0 + 1}$ = $- \frac{1}{1}$ = $- 1$.
And there you go!
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