Epsilon-delta proof

Ming1015

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Nov 7, 2020
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Hello, I would ike to ask how to prove lim x→√3 (1/√3) = ⅓. I'm struggling to assume the value of δ in terms of ε when I found out that |x-√3| = ε(3x²/x+√3) with the formula |f(x) - L| < ε when |x-x0| < δ, but I think that there's no way I could assume δ in terms of x too. May I know that is there any methods in solving it?
 
Hello, I would ike to ask how to prove lim x→√3 (1/√3) = ⅓. I'm struggling to assume the value of δ in terms of ε when I found out that |x-√3| = ε(3x²/x+√3) with the formula |f(x) - L| < ε when |x-x0| < δ, but I think that there's no way I could assume δ in terms of x too. May I know that is there any methods in solving it?
How can we possibly help. You never told us what the function is.
What is \(\bf f(x)=?\)
 
How can we possibly help. You never told us what the function is.
What is \(\bf f(x)=?\)
Hello, sir. I'm sorry, I wrote the limit wrongly. It should be lim x→√3 (1/x²) = ⅓ Just for confirmation, I'm using x0 = √3, f(x) =1/x² and then L =⅓.
 
Hello, sir. I'm sorry, I wrote the limit wrongly. It should be lim x→√3 (1/x²) = ⅓ Just for confirmation, I'm using x0 = √3, f(x) =1/x² and then L =⅓.
Start with,
\(|x-\sqrt3|<1\\-1<x-\sqrt3<1\\\sqrt3-1<x<\sqrt3+1\)
From that start you must build up the expression \(\left|\dfrac{1}{x^2}-\dfrac{1}{3}\right|=\left|\dfrac{3-x^2}{3x^2}\right|<?\)
Then pick your \(\delta=\min\left\{\dfrac{\varepsilon}{?},1\}\right\}\) and proceed.
 
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