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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https://math.berkeley.edu/~btsou/limit%20x%5E2.pdf – Braindead May 13 '15 at 13:23
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Please provide more context, and show some evidence that you really tried to solve the problem (and what you tried). This will help you receive answers that are most appropriate for your level. – Pedro M. May 13 '15 at 13:23
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Possible Duplicate of: http://math.stackexchange.com/questions/330297/prove-that-lim-x-to-2x2-4-using-epsilon-delta-definition – May 13 '15 at 13:24
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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?
mookid
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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, $$\lim_{x \to 2} x^2 = \lim_{h \to 0}(4+4h+h^2)= 4$$
Ilaya Raja S
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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) – Ilaya Raja S May 13 '15 at 15:05
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I didn't use algebra of limits and this method is purely based on the definition of limits – Ilaya Raja S May 13 '15 at 15:50