Given $f(x)=\left\{\begin{array}{1 1}\frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}, & if\;-1\leq x<0\\\frac{2x+1}{x-1}, & if\;0\leq x<1\end{array}\right.$
$\textbf{Step 1: Solving the Left Hand Limit}$:
Let $f(x)=\large \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}$
$\Rightarrow f(x)=\large \frac{(\sqrt{1+kx}-\sqrt{1-kx})(\sqrt{1+kx}+\sqrt{1-kx})}{(\sqrt{1+kx})+\sqrt{1-kx}).x}$
$\Rightarrow f(x)=\large \frac{1+kx-1+kx}{x(\sqrt{1+kx})+(\sqrt{1-kx})}$
$\Rightarrow f(x)=\large \frac{2k}{(\sqrt{1+kx})+(\sqrt{1-kx})}$
Since the limit is $0$, substituting, $x=0$, we get
$\Rightarrow f(x)=\large \frac{2k}{(\sqrt{1+k(0)})+(\sqrt{1-k(0)})}=\frac{2k}{2}\normalsize =k.$
$\textbf{Step 2: Solving the Right Hand Limit}$:
Let $f(x) = \large \frac{2x+1}{x-1}$
Since the limit is $0$, substituting, $x=0$, we get
$\Rightarrow f(x)=\large \frac{2(0)+1}{0-1}\normalsize =-1.$
$\textbf{Step 3: Solving for k}$:
Since $f(x)$ is continuous at $0$, Left Hand Limit = Right Hand Limit.
$\Rightarrow k = -1$.