how to prove limit exits using sequential limit definition

Question

Let $f:(-\infty, 1) \cup (1,\infty) \to \mathbb{R}$ be defined by $f(x)=\frac{x^3-1}{x-1}$ . Prove that $\lim _{x\to 1}f(x)=3$

i am proving this problem by limit of the definition

$|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)=4|x-1|<\epsilon $ as i choose $|x-1|<\delta=\epsilon/4$

how to prove this by sequence of limit

ie how to prove by if $x_n \to 1$ then $f(x_n)\to 3$

Answer 1

Let it be given that $x_n \to 1$. This means you can make $|x_n-1| < \epsilon$, taking $n$ big enough.

We want to show $f(x_n) \to 3$, so we want to show that we can make |$f(x_n)-3|<\epsilon$, by taking large enough $n$.

But you know that $|f(x_n)-3| = |x_n-1||x_n+2|$. The first factor can be made $ < \epsilon$ by assumption, and the second one can be made arbitrarily close to $3$ for a large enough $n$.

All that's left is writing up all the details.

Answer 2

You already know that $f(x)-3=x^{2}+x-2$. If $x_n \to 1$ then $x_n^{2} \to 1$ so $f(x)-3 \to 1+1-2=0$ and $f(x) \to 3$.