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?

Question

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!

Answer 1

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 $b=1$.

Answer 2

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

$$\lim_{x\to 1}\frac{(2x^2-1)^{1/3}-x^{1/2}}{x-1}=\lim_{x\to 1}\left(\frac13(2x^2-1)^{-2/3}4x-\frac12 x^{-1/2}\right)=\frac56$$

Answer 3

$$ \begin{align} \lim_{x\to1}\frac{\left(2x^2-1\right)^{\frac13}-x^{\frac12}}{x-1} &=\lim_{x\to0}\frac{\left(2x^2+4x+1\right)^{\frac13}-(x+1)^{\frac12}}{x}\tag{1}\\ &=\lim_{x\to0}\frac{\left(2x^2+4x+1\right)^{\frac13}-1}{x}-\lim_{x\to0}\frac{(x+1)^{\frac12}-1}{x}\tag{2}\\ &=\lim_{x\to0}\frac{2x^2+4x}{x\left(\left(2x^2+4x+1\right)^{\frac23}+\left(2x^2+4x+1\right)^{\frac13}+1\right)}\tag{3}\\ &-\lim_{x\to0}\frac{x}{x\left((x+1)^{\frac12}+1\right)}\tag{4}\\ &=\lim_{x\to0}\frac{2x+4}{\left(2x^2+4x+1\right)^{\frac23}+\left(2x^2+4x+1\right)^{\frac13}+1}\tag{5}\\ &-\lim_{x\to0}\frac{1}{(x+1)^{\frac12}+1}\tag{6}\\ &=\frac43-\frac12\tag{7}\\[3pt] &=\frac56\tag{8} \end{align} $$ 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