If $\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\frac{\sqrt{1+\mathrm{kx}}-\sqrt{1-\mathrm{kx}}}{\mathrm{x}} -1<\mathrm{x}<0\\ 2{\mathrm{x}}^2+3\mathrm{x}-2 0\le \mathrm{x}<1\end{array}\right.$ is continuous at x = 0 then K = ________

The given function is $\mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\frac{\sqrt{1+\mathrm{kx}}-\sqrt{1-\mathrm{kx}}}{\mathrm{x}} -1<\mathrm{x}<0\\ 2{\mathrm{x}}^2+3\mathrm{x}-2 0\le \mathrm{x}<1\end{array}\right.$

Given that the function is continuous at $x=0$

Hence, $\underset{x\to 0}{\lim }f\left(x\right)=f\left(0\right)$

$\begin{array}{rcl}2{\left(0\right)}^2+3\left(0\right)-2& =& \underset{x\to 0}{\lim }\frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}\\ -2& =& \underset{x\to 0}{\lim }\frac{1+kx-1+kx}{x\left(\sqrt{1+kx}+\sqrt{1-kx}\right)}\\ -2& =& \underset{x\to 0}{\lim }\frac{2k}{\left(\sqrt{1+kx}+\sqrt{1-kx}\right)}\\ -2& =& k\end{array}$