Calcul-(us)-ator

This is my solution.Yours may be different.

Firstly,we recall and use some $Trigonometrical$ and $Logarithmic$ Series namely $e^x$ $=$ $1$ $+$ $x$ $+$ $\frac{x^2}{2!}$ $+$ $\frac{x^3}{3!}$ $+$ ... , then we have $sin$ $x$ $=$ $x$ $-$ $\frac{x^3}{3!}$ $+$ $\frac{x^5}{5!}$ $-$ ...( $x$ is in $radians$ ) , then $log_e(1-x)$ $=$ $-$ $x$ $-$ $x^2/2$ $-$ $x^3/3$ $-$ ...

For the second limit we have $sin$ $x$ which has been already mentioned, $\frac{1}{2}$ $log_e$ $\frac{1+x}{1-x}$ $=$ $x$ $+$ $\frac{x^3}{3}$ $+$ $\frac{x^5}{5}$ $+$ ...

Finally using these special series on evaluating the limits separately (both limits are undetermined of the form $0/0$ ) by cancelling $x$ and applying $x$ $=$ $0$ after cancelling we arrive at $\frac{13}{60}$ $+$ $\frac{-2}{3}$ $=$ $-0.45$ .

Edit: 4/7/17 hey now i now another solution we could simply differentiate both limits (L Hopital 's rules since it is of 0/0 form that 's easier instead of remembering all those taylor series.It's far easier to use this approach.