Determine that the function $f(x)=\dfrac{x^2+x-2}{x-1}$ is continous in $x_0=0$; $x_0\in \mathbb{R}$

Question

Determine that the function $f(x)=\dfrac{x^2+x-2}{x-1}$ is continous in $x_0=0$; $x_0\in\mathbb{R}$ (I only need to show continuity for $x_0=0$)

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Let $\varepsilon>0$; ${\mid x-x_0\mid}<\delta = {\mid x-0 \mid}<\delta$ \begin{align} {\mid f(x)-f(x_0)\mid} &= {\mid\frac{x^2+x-2}{x-1}-\frac{0^2+0-2}{0-1}\mid}\\ &={\mid\frac{x^2+x-2}{x-1}-2\mid}\\ &={\mid\frac{x^2-x}{x-1}\mid}\\ &={\mid x\mid}<\delta\\ \end{align} Let $\delta = \varepsilon$, then if $|x-0|<\delta$ then $|f(x)-f(0)| = |x| <\epsilon$

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Is there anything that should be changed?

Answer 1

Your idea is correct and the choice of $\delta = \epsilon$ is right. I would make some changes to the writing to make it more readable to those who are not as familiar with the ideas as you are.

For example your last line of proof may be altered as :

Let $\epsilon = \delta$, then if $|x-0|<\delta$, then $|f(x)-f(0)| = |x| <\epsilon$

Answer 2

It might be just a typo, but you wrote $\delta =: \varepsilon$, saying "let's define $\varepsilon$ to be equal to $\delta$." This would be a mistake, since $\varepsilon$ is arbitrary but fixed in the beginning of a proof and it is $\delta$ that should be defined for a particular $\varepsilon$.

Otherwise, your work looks solid. It could be simplified if you notice that $x^2+x-2 = (x+2)(x-1)$, so $f(x) = x + 2$ when $x\neq 1$ (and undefined at $x = 1$).