prove that $\lim _{x\to 1}f(x)=3$ if $f(x)=\frac{x^3-1}{x-1}$

Question

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

Answer 1

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. $|f(x)-3|<\epsilon$.

Answer 2

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.

Answer 3

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

$$\lim _{x\to 1}\frac{x^3-1}{x-1}=\lim _{x\to 1}\,\frac{(x-1)(x^2+x+1)}{x-1}=\lim _{x\to 1}\,(x^2+x+1)=1+1+1=3$$

Answer 4

$x^3-1=(x-1)(x^2+x+1)$.

Hence $\lim_{x\to1}\frac{x^3-1}{x-1}=\lim_{x\to1}x^2+x+1=1^2+1+1=3$.

Answer 5

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 $\lim_{x\to 1} \frac{x^3-1}{x-1} = 3$.

Answer 6

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}