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Given the function difined by $f(x)=x^2-2x+1$

Fin real $k$ such that:

For every $x \in(2,4)$ : $|f(x)-4|\leq k|x-3|$

Fin $\delta$ such that :

$|x-3|<\delta$ => |$f(x)-4|<\epsilon$

The deduce the limit of $f(x)$ in the point $3$

$|x^2-2x+1-4|=|x^2-2x-3|=|x-3||x+1|$

$x \in (2,4) |x+1|\leq 5$ so $k=5$

And so :

$|x-3|<\epsilon$ and $|x+1|<5$ so

$\delta=\inf(\frac{\epsilon}{5};1)$

Does this have any relation with MVT (mean value theorem)

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  • $\begingroup$ Any hints ????? $\endgroup$ – user233658 Mar 12 '16 at 22:05

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