using the Epsilon Delta definition of a limit and not what was taught in calculu

Question

using the Epsilon Delta definition of a limit and not what was taught in calculus prove the following with: For all epsilon>0 there exist delta>0 s.t for all x, 0<|x-x_0|< delta implies |f(x)-L|< Epsilon.

I can solve this problem using calculus,so please do not answer if you don't know how to do this other than with calculus.

I know the answer for this is when delta = Epsilon but I don't know how to do a nice write up for this problem so that I could use other problems like this to solve for other limits = infinity.

i need help in the write up and the thought process. thank you in advance.

Explanation / Answer

To prove that limit is infinity then for all ,M>0, delta (d)>0 there exist z so that

if |z-2|<d then, |f(z)|>M

Let, z=2-d so we ahve

2/(2-z)=2/d,d>0

By Archimidean property there exist,p, so that:2/p>M

If p>d then choose :z=d/2

Otherwise we are done.

Hence proved

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