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# Prove the statement using the $\varepsilon$, $\delta$ definition of a limit.$\displaystyle \lim_{x \to 2} (x^2 - 4x + 5) = 1$

## Given $\epsilon>0,$ we need $\delta>0$ such that if $0<|x-2|<\delta$, then $\left|\left(x^{2}-4 x+5\right)-1\right|<\varepsilon \Leftrightarrow\left|x^{2}-4 x+4\right|<\varepsilon \Leftrightarrow$$\left|(x-2)^{2}\right|<c .$ So take $\delta=\sqrt{\epsilon} .$ Then $0<|x-2|<\delta \Leftrightarrow|x-2|<\sqrt{\epsilon} \Leftrightarrow\left|(x-2)^{2}\right|<c .$ Thus,$\lim _{x \rightarrow 2}\left(x^{2}-4 x+5\right)=1$ by the definition of a limit.

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This is problem number twenty nine of this tour. Calculus, eighth edition, Section two point four. Prove this statement using the AB signed out a definition of probable over limit. The limit is experts is too of the function X squared when this works was five is equal to one. The Epsilon Delta definition of limit is that for any absolute grade in zero, there must be in delta greater than zero such that if the Sioux Valley of the difference between X and A is less than daughter, then the absolutely of the different between the function and the limit is less than Absalon. Our first step will be to use a second inequality. Here, use our function X squared minus four X plus five. So we just play this into the second equality. Or here we're going to subtract R l. How is the limit? The limit is equal to one. And then recent this Liston Absalon. If we simplify the inside combined the plus five in the minus one, it leaves us with excluding menace for explosives. For if we factor, this is a quadratic that could be written in the form X wings to quantity squared because this is a perfect square. If we fact is out foil, we'LL get X squared minus works because I think the simplest thing to is making for and then they get up to ten thing attitude is positive for and our final step will be to take the square root of four sites now working with the first inequality. The value for our problem is the value that exit purchase which is too. And this is our delta. I noticed that the last inequality that we found for Absalom matches very similarly to this adulterer and equality. And as such, we can choose Doctor to be equal to square root of Absalon which gives us a solution on such that we found it. Find any Delta we want provided. And Absalon So we choose any insulin we want. We get our we know any Delta from that ups on that matches thes two inequalities. And as such, we have a proven this statement limit I'm using this app signed off the definition of the limit

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