<s>[INST] prove that $\lim _{x\to 1}f(x)=3$ if $f(x)=\frac{x^3-1}{x-1}$ Prove that $\lim _{x\to 1}f(x)=3$ where $f:(0,\infty)\to \mathbb{R}$ is given by $f(x)=\frac{x^3-1}{x-1}$.
I proved by definition of the limit 
$$|f(x)-3|=\left|\frac{x^3-1}{x-1}-3\right|=|x^2+x-2|=|x-1||x+2|\leq |x-1|(|x|+2)$$
how to processed from this
 [/INST] By definition of the limit 
$$f(x)-3=\frac{x^3-1}{x-1}-3=x^2+x-2=(x-1)^2+3(x-1)$$
\begin{align}
|f(x)-3|
&= |(x-1)^2+3(x-1)| \\
&= |x-1|^2+3|x-1| \\
&<\delta^2+3\delta \\
&= \varepsilon
\end{align}
</s>