why $e^{(x/(x - 1))}-1>0$ for all $x$ just to the left of $x = 0,$

Question

why $e^{(x/(x - 1))}-1>0$ for all $x$ just to the left of $x = 0$? Detailed process picture

Answer 1

If $x<0,$

then $x-1<0,$

so $t=\dfrac x{x-1}>0,$

so $e^{t}>1,$

so $e^t-1>0$.