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
 A: $x^3-1=(x-1)(x^2+x+1)$.
Hence $\lim_{x\to1}\frac{x^3-1}{x-1}=\lim_{x\to1}x^2+x+1=1^2+1+1=3$.
A: Hint:
From
$$x^2+x-2=(x-1)(x+2)$$ you deduce that the expression will tend to zero thanks to the first factor, and in a finite neighborhood, say $[0,2]$ the second factor remains bounded (does not exceed $4$).
Hence,
$$|x^2+x-2|\le 4|x-1|.$$
This allows you to find the $\epsilon-\delta$ combinations.
A: As an alternative by definition of derivative with $g(x)=x^3\implies g'(x)=3x^2$
$$\lim _{x\to 1}\frac{x^3-1}{x-1}=g'(1)=3$$
or by factorization
$$\lim _{x\to 1}\frac{x^3-1}{x-1}=\lim _{x\to 1}\,\frac{(x-1)(x^2+x+1)}{x-1}=\lim _{x\to 1}\,(x^2+x+1)=1+1+1=3$$
A: Let $\varepsilon > 0$. We have
$$\left|\frac{x^3-1}{x-1} - 3\right| = \left|\frac{x^3-3x+2}{x-1}\right| = \left|x^2+x-2\right| = |x-1||x+2| \le |x-1|(|x-1| + 3)$$
Solving the quadratic inequation $y^2+3y - \varepsilon < 0$ gives $y \in \left\langle \frac{-3-\sqrt{9+4\varepsilon}}2, \frac{-3+\sqrt{9+4\varepsilon}}2\right\rangle$.
Therefore if we pick $\delta = \frac{-3+\sqrt{9+4\varepsilon}}2$, then $|x-1| < \delta$ implies 
$$\left|\frac{x^3-1}{x-1} - 3\right| \le |x-1|^2 + 3|x-1| < \varepsilon$$
We conclude $\lim_{x\to 1} \frac{x^3-1}{x-1} = 3$.
A: Limit of a product is product of the limits. Since the limit of $|x-1|$ is $0$ and the limit of $|x|+2$ is 3, the limit of the product is $0$. For an $\epsilon- \delta$ argument let $\epsilon >0$ and choose $\delta \in (0,1)$ such that $\delta <\epsilon /4$. Then $|x-1| <\delta$ implies $|x-1| <\epsilon /4$ and $|x|+2<4$ so the product is less than $(\epsilon /4) (4)=\epsilon$, i.e. $|f(x)-3|<\epsilon$.
A: 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}
