Suppose $\epsilon> 0$ and $\displaystyle\delta = \min \left\{1,\frac{\epsilon}{10}\right\}$

If $0 < |x -1 |< \delta$, does this imply $|x^{2}-1|<\epsilon$?

I have verified that for $\epsilon=20$, $\delta=1$ and $x=1.9$, it is true.

I have also verified that for $\epsilon=5$, $\displaystyle\delta=\frac{1}{2}$ and $x=1.2$, it is true.

I am having problems proving it.

Now $|x^{2}-1|<\epsilon$ is $|x-1|\cdot|x+1|<\epsilon$

And I know $\displaystyle2<|x+1|<\frac{\epsilon}{10}+2$, or


But not sure where to go next.


to solve this type of problem you have to find relationship between $\delta$ and $\epsilon$ in worst case. in other word you have to find an upper bound for $\delta$ that satisfies expression related to $\epsilon$.

for example we have to start from $|f(x)-L|<\epsilon$ and conclude something like $|x-a||g(x)|<\epsilon$. then by finding an upper bound on $|g(x)|$ like $M$ we can conclude that $\delta<min\{\frac{\epsilon}{M},\delta_1\}$ where $\delta_1$ appears in some cases that function $f(x)$ is not defined in the specified interval. so $\delta_1$ is value that $f(x)$ is well defined at $(a-\delta_1,a+\delta_1)$

to reach this goal we can write: $$|x^{2}-1|<\epsilon \Rightarrow |x-1|\cdot|x+1|<\epsilon$$ we see that since $x \to 1$ therefore an upper bound on $|x+1|$ is $3$. so it can be concluded that $|x-1|\cdot|x+1|<3\epsilon$. so in order to ensure $|x-1|<\delta$ yields $|x^2-1|<\epsilon$ we have $\delta<\frac{\epsilon}{3}$.

so we can conclude that $\delta = \min\{1,\frac{\epsilon}{10}\}$ gives us $|x^{2}-1|<\epsilon$ according to transitive property of Real numbers.


$0\leq |x-1||x+1|<\delta(\delta+2)$ because if $x\geq 1$ then $x-1<\delta$ implies $x+1<2+\delta$ and if $x<1$ then $x+1<2$ and so $|x+1|<2+\delta$ for $x\geq-1$. If $x<-1$ then $|x-1|=1-x<\delta$ implies $|x+1|=-x-1<\delta-2<\delta+2$. If $\epsilon\leq 10$ then $\delta=\frac{\epsilon}{10}$ and thus


where the last inequality holds for $\epsilon\leq 80$.

Finally if $\epsilon>10$ then $0\leq |x-1||x+1|<\delta(\delta+2)=3<\epsilon $.


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