<s>[INST] 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!
 [/INST] It mostly depends on what you're allowed to use. In order to do it without recurring to derivatives, I'd split it into two parts:
\begin{align}
\lim _{x\to 1}\biggl(\frac{(2x^2-1)^{1/3}-x^{1/2}}{x-1}\biggr)
&=\lim _{x\to 1}\biggl(\frac{(2x^2-1)^{1/3}-1+1-x^{1/2}}{x-1}\biggr)
\\[6px]
&=\lim_{x\to 1}\left(\frac{(2x^2-1)^{1/3}-1}{x-1}-
\frac{x^{1/2}-1}{x-1}\right) \\[6px]
&=\lim_{x\to 1}\frac{(2x^2-1)^{1/3}-1}{x-1}-
\lim_{x\to 1}\frac{x^{1/2}-1}{x-1}
\end{align}
provided both limits in the last expression exist (they do). The second one is easy: multiply by $x^{1/2}+1$ to get
$$
\lim_{x\to 1}\frac{x^{1/2}-1}{x-1}=
\lim_{x\to 1}\frac{x-1}{(x-1)(x^{1/2}+1)}=
\lim_{x\to 1}\frac{1}{x^{1/2}+1}=\frac{1}{2}
$$
The first one can be treated in a similar way recalling that
$$
a^3-b^3=(a-b)(a^2+ab+b^2)
$$
where $a=(2x^2-1)^{1/3}$ and $b=1$.
</s>