# Prove a limit, going back to definition

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!

$$|x^2 - 4| = |x-2||x+2|$$ 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 $\epsilon$?
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, $$\lim_{x \to 2} x^2 = \lim_{h \to 0}(4+4h+h^2)= 4$$