Evaluating $\lim _{x\to 1}\left(\frac{\left(2x^2-1\right)^{\frac{1}{3}}-x^{\frac{1}{2}}}{x-1}\right)$ Without L'Hopital or Calculus? What is: $\lim _{x\to 1}\left(\frac{\left(2x^2-1\right)^{\frac{1}{3}}-x^{\frac{1}{2}}}{x-1}\right)$?
Thanks in advance
Much appreciated!

$$
\begin{align}
\lim_{x\to1}\frac{\left(2x^2-1\right)^{\frac13}-x^{\frac12}}{x-1}
&=\lim_{x\to0}\frac{\left(2x^2+4x+1\right)^{\frac13}-(x+1)^{\frac12}}{x}\tag{1}\\
&=\lim_{x\to0}\frac{\left(2x^2+4x+1\right)^{\frac13}-1}{x}-\lim_{x\to0}\frac{(x+1)^{\frac12}-1}{x}\tag{2}\\
&=\lim_{x\to0}\frac{2x^2+4x}{x\left(\left(2x^2+4x+1\right)^{\frac23}+\left(2x^2+4x+1\right)^{\frac13}+1\right)}\tag{3}\\
&-\lim_{x\to0}\frac{x}{x\left((x+1)^{\frac12}+1\right)}\tag{4}\\
&=\lim_{x\to0}\frac{2x+4}{\left(2x^2+4x+1\right)^{\frac23}+\left(2x^2+4x+1\right)^{\frac13}+1}\tag{5}\\
&-\lim_{x\to0}\frac{1}{(x+1)^{\frac12}+1}\tag{6}\\
&=\frac43-\frac12\tag{7}\\[3pt]
&=\frac56\tag{8}
\end{align}
$$
Explanation:
$(1)$: substitute $x\mapsto x+1$
$(2)$: subtract and add $1$ to the numerator so that we can split the limit into two
$(3)$: multiply and divide by $\left(2x^2+4x+1\right)^{\frac23}+\left(2x^2+4x+1\right)^{\frac13}+1$
$(4)$: multiply and divide by $(x+1)^{\frac12}+1$
$(5)$: cancel $x$ in numerator and denominator
$(6)$: cancel $x$ in numerator and denominator
$(7)$: evaluate $(5)$ and $(6)$ at $x=0$
$(8)$: simplify
