How to calculate limit in a quotient function How to calculate the limit of the following function:

$$\lim_{x \rightarrow 1}\frac{\sqrt[3]{2x^2-1} -\sqrt[2]{x} }{x-1}$$

I've tried to use the following formula: $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ without any success. Is there anything that I'm missing?

I can't use L'Hôpital rule.
Basic limit laws and standard limit $$\lim_{x\to a} \frac{x^n-a^n}{x-a}=na^{n-1}\tag{1}$$ suffice here.

We have
\begin{align*}
L&amp;=\lim_{x\to 1}\frac{\sqrt[3]{2x^2-1}-\sqrt{x}}{x-1}\\
&amp;=\lim_{x\to 1}\frac{\sqrt[3]{2x^2-1}-1}{x-1}-\frac{x^{1/2}-1^{1/2}}{x-1}\\
&amp;=\lim_{x\to 1}\frac{\sqrt[3]{2x^2-1}-1}{2x^2-1-1}\cdot 2(x+1)-\frac{1}{2}\\
&amp;=4\lim_{t\to 1}\frac{t^{1/3}-1^{1/3}}{t-1}-\frac{1}{2}\text{ (putting }t=2x^2-1) \\
&amp;=4\cdot\frac{1}{3}-\frac{1}{2}\\
&amp;=\frac{5}{6}
\end{align*}
With $x=y+1$, you get
$$\frac{\sqrt[3]{2x^2-1}-\sqrt{x}}{x-1} = \frac{\sqrt[3]{2(y+1)^2-1}-\sqrt{y+1}}{y}$$

You have to compute the limit of this quantity when $y$ tends to $0$.

Now 
$$\frac{\sqrt[3]{2(y+1)^2-1}-\sqrt{y+1}}{y} = \frac{\sqrt[3]{2y^2 + 4y +1}-\sqrt{y+1}}{y} = \frac{1+ \frac{1}{3}(2y^2+4y) + o(y)-1- \frac{1}{2}y + o(y)}{y} $$

i.e. 
$$\frac{\sqrt[3]{2(y+1)^2-1}-\sqrt{y+1}}{y} = \frac{5}{6} + o(1)$$

So the limit is equal to 
$$\frac{5}{6}$$