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Here are some examples of proofs I struggle with:

  1. Consider the sequence {$15^2+2√+3$} $\lim\limits_{→∞} 15^2+2√+3$=_____

  2. Let $()=2^2−4+5$. Use the − definition to prove that $\lim\limits_{ \to -1} 2^2−4+5≠8$.

I'm struggling with how to solve these proofs in general, my textbook hasn't been great.

Mark
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Hillel L.
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  • https://math.stackexchange.com/questions/1745550/logic-behind-epsilon-delta-proofs-single-variable-calculus?rq=1 – Mark Nov 30 '22 at 22:23
  • You are allowed to cheat to get inspiration. For 2) we know $2x^2-4x +5$ is continuous so we know that $\lim_{x\to -1} 2x^2-4x + 5$ is actually $2(-1)^2 -4(-1) +5=11\ne 8$. So we know as $x$ gets close to $-1$, $f(x)$ doesn't get close to $8$. So show that for some $\epsilon$ there is a $\delta$ small enough that $-1-\delta<x< -1+\delta$ then $f(x) > 8 + \epsilon$. This shouldn't be hard as if $\delta$ is small enough $f(x)\approx 11$. If say $\delta=\frac 14$ for and $\epsilon=1$ then for $-1.25<x<-0.75$ we have $f(x) > 9 =8+1 > 8+\epsilon$. That proves the limit is not $8$. – fleablood Dec 01 '22 at 00:32
  • It should be intuitively obvious that as $n\to \infty$ that $15n^2\to \infty$ and $\sqrt n\to \infty$ so $\lim_{n\to \infty} 15n^2+ \sqrt n + 3=\infty$. The proof needs to show that for any $K$ no matter how big there is an $N$ so that for all $n > N$ we have $15n^2 + \sqrt n + 3 > K$. Well, there's no bonus being close. If $n > N> 1$ and $N \ge K$ then $15n^2 + \sqrt n + 3 > 15n^2 > n^2 > n > N \ge K$ and that's it we are done. – fleablood Dec 01 '22 at 00:38
  • Does this answer your question? Does there exist an epsilon-delta proof of the limit for all polynomials? –  Oct 07 '23 at 23:14

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