How is this limit being solved? I can't grasp it

Question

I am going over limits for my finals as I notice this example in my schoolbook discribing limits of the undefined form $0\over0$ in the shape of an irrational fraction.

$$\lim\limits_{x \to 1} {\sqrt[3]{x}-1\over\sqrt{2x-1}-1} = {0\over0} = \lim\limits_{x \to 1} {(\sqrt[3]{x}-1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)(\sqrt{2x-1}+1)\over(\sqrt{2x-1}-1)(\sqrt{2x-1}+1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)}$$

I really don't get what is going on here. I know that you are supposed to multiply both the nominator and the denominator by the added value of either. But it appears that here it has been mutliplied by both? And where does this term $(\sqrt[3]{x^2}+\sqrt[3]{x}+1)$ come from?

There is no worked out example, except for the solution after this which is $1\over3$ but thats it.

Could anyone please explain this to me? I would be really gratefull! Also sorry if I didn't use the correct terms here and there, not a native english speaker.

Thanks in advance. Cheers, Michiel

Answer 1

HINT: $(a-b)(a^2+ab+b^2)=a^3-b^3$ and $(a+b)(a-b)=a^2-b^2$.

Answer 2

$${\sqrt[3]{x}-1\over\sqrt{2x-1}-1} = \frac{\sqrt[3]{x}-1}{\sqrt{2x-1}-1}\cdot \frac{(\sqrt{2x-1}+1)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}{(\sqrt{2x-1}+1)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}$$

$$=\frac{\left((\sqrt[3]{x}-1)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)\right)(\sqrt{2x-1}+1)}{\left((\sqrt{2x-1})^2-1^2\right)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}$$

$$=\frac{(x-1)(\sqrt{2x-1}+1)}{2(x-1)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}=\frac{\sqrt{2x-1}+1}{2\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}$$

$$\stackrel{x\to 1}\to \frac{\sqrt{2(1)-1}+1}{2\left(\sqrt[3]{1^2}+\sqrt[3]{1}+1\right)}=\frac{1}{3}$$

Answer 3

Where does $\sqrt[3]{x^2}+\sqrt[3]{x}+1$ comes from? The identity you can use is $$ a^3-1=(a-1)(a^2+a+1) $$ with $a=\sqrt[3]{x}$; so, multipliying numerator and denominator by $\sqrt[3]{x^2}+\sqrt[3]{x}+1$ you get rid of part of the indetermination, together with the fact that, multiplying numerator and denominator by $\sqrt{2x-1}+1$ the denominator becomes $(2x-1)-1=2(x-1)$, so the factors $x-1$ cancel each other.

Let me show a different way for computing the limit. With a bit of experience, you can split this into $$ \lim_{x\to1}\frac{\sqrt[3]{x}-1}{x-1} \cdot \lim_{x\to1}\frac{x-1}{\sqrt{2x-1}-1} $$ (provided both limits exist, and this is where experience gets in).

The first limit can be done by a simple substitution, $x=t^3$: $$ \lim_{x\to1}\frac{\sqrt[3]{x}-1}{x-1}= \lim_{t\to1}\frac{t-1}{t^3-1}= \lim_{t\to1}\frac{1}{t^2+t+1}=\frac{1}{3} $$ Similarly, for the second limit you can set $\sqrt{2x-1}-1=t$, so $2x-1=t^2+2t+1$, $x=(t^2+2t+2)/2$ and $x-1=(t^2+2t)/2$; therefore $$ \lim_{x\to1}\frac{x-1}{\sqrt{2x-1}-1}= \frac{1}{2}\lim_{t\to0}\frac{t^2+2t}{t}= \frac{1}{2}\lim_{t\to0}(t+2)=1 $$

To do it with Taylor expansions, substitute $x=t+1$, so the limit becomes $$ \lim_{t\to0}\frac{\sqrt[3]{1+t}-1}{\sqrt{1+2t}-1}= \lim_{t\to0}\frac{1+\frac{1}{3}t+o(t)-1}{1+\frac{1}{2}(2t)-1}= \lim_{t\to0}\frac{\frac{1}{3}t+o(t)}{t+o(t)}=\frac{1}{3} $$