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!
 A: $$
\begin{align}
\lim_{x\to1}\frac{\left(2x^2-1\right)^{\frac13}-x^{\frac12}}{x-1}
&=\lim_{x\to0}\frac{\left(2x^2+4x+1\right)^{\frac13}-(x+1)^{\frac12}}{x}\tag{1}\\
&=\lim_{x\to0}\frac{\left(2x^2+4x+1\right)^{\frac13}-1}{x}-\lim_{x\to0}\frac{(x+1)^{\frac12}-1}{x}\tag{2}\\
&=\lim_{x\to0}\frac{2x^2+4x}{x\left(\left(2x^2+4x+1\right)^{\frac23}+\left(2x^2+4x+1\right)^{\frac13}+1\right)}\tag{3}\\
&-\lim_{x\to0}\frac{x}{x\left((x+1)^{\frac12}+1\right)}\tag{4}\\
&=\lim_{x\to0}\frac{2x+4}{\left(2x^2+4x+1\right)^{\frac23}+\left(2x^2+4x+1\right)^{\frac13}+1}\tag{5}\\
&-\lim_{x\to0}\frac{1}{(x+1)^{\frac12}+1}\tag{6}\\
&=\frac43-\frac12\tag{7}\\[3pt]
&=\frac56\tag{8}
\end{align}
$$
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
A: We can expand the terms in the numerator as
$$(2x^2-1)^{1/3}=1+\frac43(x-1)+O(x-1)^2 \tag 1$$
and
$$x^{1/2}=1+\frac12(x-1)+O(x-1)^2 \tag2$$
Subtracting $(2)$ from $(1)$ and passing to the limit we find
$$\lim_{x\to 1}\frac{(2x^2-1)^{1/3}-x^{1/2}}{x-1}=\frac43-\frac12=\frac56$$
To verify, L'Hospital's Rule comes to the rescue.  We have
$$\lim_{x\to 1}\frac{(2x^2-1)^{1/3}-x^{1/2}}{x-1}=\lim_{x\to 1}\left(\frac13(2x^2-1)^{-2/3}4x-\frac12 x^{-1/2}\right)=\frac56$$
A: 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 $b=1$.
