How do you solve $\lim_{n\to \infty}{n[(1+\frac{c}{n})^{n}-(1-\frac{c}{n})^{-n}]}$ without L’Hopital's Rule? I know that solution is: $-c^2e^c$ 
$$\lim\nolimits_{n\to \infty}{n\left[\left(1+\frac{c}{n}\right)^{n}-\left(1-\frac{c}{n}\right)^{-n}\right]}$$
Hint: common factor to $\left(1+\frac{c}{n}\right)^{n}$ and then to use approaches
 A: As a big hint, assuming WLOG that $c > 0$, we have as $n \to \infty$
$$\left(1+\frac{c}{n}\right)^{n}\uparrow e^c  \downarrow\left(1-\frac{c}{n}\right)^{-n},$$
and for all $n > c$
$$0 \leqslant n\left[\left(1-\frac{c}{n}\right)^{-n}-\left(1+\frac{c}{n}\right)^{n}\right] = n\left(1-\frac{c}{n}\right)^{-n}\left[1 -  \left(1-\frac{c^2}{n^2}\right)^{n}\right] \leqslant \ldots$$
See if you can finish by first applying Bernoulli's inequality to the last term on the RHS and then applying the squeeze theorem.
A: Hint: $e^x = \lim_{n\to+\infty} \left[1+\frac{x}{n}\right]^n$
A: $$
\lim\limits_{n\to\infty} \left[ n\left( 1 + \frac{c}{n} \right)^n - n\left( 1 - \frac{c}{n} \right)^{-n}\right] = 
\lim\limits_{n\to\infty} \left[ (n + c)\left( 1 + \frac{c}{n} \right)^{n-1} - (n - c)\left( 1 - \frac{c}{n} \right)^{-n - 1}\right] = \\
\lim\limits_{n\to\infty} \left[ (n + c)\underbrace{\left( 1 + \frac{c}{n} \right)^{-1}}_{1}\underbrace{\left( 1 + \frac{c}{n} \right)^{n}}_{e^c} - (n - c)\underbrace{\left( 1 - \frac{c}{n} \right)^{-1}}_{1}\underbrace{\left( 1 - \frac{c}{n} \right)^{-n}}_{e^c}\right] = 2\cdot c \cdot e^c\\
$$
