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I have to use only the definition of limits (ie I can't use algebra of limits) to prove the following:

$$\lim_{x \to 2} x^2 = 4$$

I can't think of what to use as an arbitrary constant, or how to start this, thanks.

full proofs would be brilliant!

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You want to make

$$ |x^2 - 4| = |x-2||x+2| $$ small, and $|x-2|$ is allowed to be small itself. So, you just need to restrain the domain of $x$ such as $$ |x+2| $$ is not too big. Say $|x+2| < 5$. What are the implications for the interval of $x$? can $x$ be close to 2 in this interval? What is then the choice of $\delta$ in terms of $\epsilon$?

Can you take it from here?

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Use h method First substitute x with $2+h$ where $h \to 0^+$ ( is a small positive quantity) $$\lim_{h \to 0} (h+2)^2$$ Then expanding it comes out to be. $$ \lim_{h \to 0}(4+4h+h^2) $$ Then as $h$ is very small ignore $4h$ and $h^2$ we get, $$\lim_{x \to 2} x^2 = \lim_{h \to 0}(4+4h+h^2)= 4$$

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  • $\begingroup$ I recommend using this method in future applications also. Don't treat this as a solution but as a good approach for most questions (it's very basic too) $\endgroup$ – Ilaya Raja S May 13 '15 at 15:05
  • $\begingroup$ although this works it doesnt use the definition of limits! $\endgroup$ – Lauren Bathers May 13 '15 at 15:48
  • $\begingroup$ I didn't use algebra of limits and this method is purely based on the definition of limits $\endgroup$ – Ilaya Raja S May 13 '15 at 15:50

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