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$$\lim_{x\to a} x^4 = L$$ for some arbitrary a

Picking $a$ to be 2, we get:

$$\lim_{x\to 2} x^4 = 16$$

To show that is the limit I tried doing the epsilon-delta definition of a limit to show how to find a $δ$ such that $|f(x) - L| < \epsilon $ for all x satisfying $0 < |x-a| < δ$

And here's how I attempted it:

$\forall ε>0, \exists δ>0$, such that for all x, if $0<|x-2|<δ$ then $|x^4 - 16| < ε$

$$|x^4 - 16| < ε$$ $$|(x-2)(x+2)(x^2+4)| < ε$$

$$δ: |x-2| < δ$$ I picked $δ$ to be 1, then,

$$|x-2| < 1 \Rightarrow 1 < x < 3 \Rightarrow 3 < x + 2 < 5 \Rightarrow 7 < x^2 + 4 < 9$$

so,

$$|(x-2)(x+2)(x^2+4)| < |x-2|*9 < ε \Rightarrow |x-2| < \frac{ε}{9}$$

therefore,

$$δ: min\lbrace1, \frac{ε}{9}\rbrace$$

I was wondering if what I did was correct and if it isn't can someone show me where I might of messed up.

Sc4r
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    x^2 + 4 < 13 and not 9, and /x + 2/ < 5. So delta = min{1, e/65} – DeepSea Mar 03 '14 at 02:20
  • I think this sites will be helpful in the future: http://www.ocf.berkeley.edu/~yosenl/math/epsilon-delta.pdf , http://educ.jmu.edu/~taalmala/235_2000post/235epsilondelta.pdf , http://mathworld.wolfram.com/Epsilon-DeltaProof.html – Ana Galois Mar 03 '14 at 02:58

2 Answers2

1

Let's start at d = 1 ( d = delta ) : 1 < x < 3 ==> 1 < x^2 < 9 ==> 5 < x^2 + 4 < 13 ( yours is 9 ). Next, 3 < x + 2 < 5 ==> /x + 2/ < 5. Finally: /x^2 - 16/ = /(x - 2)(x + 2)(x^2 + 4)/ < 5*13*/x - 2/ = 65*/x - 2/. We need that 65*/x - 2/ < e ( e = epsilon ) ==> /x - 2/ < e/65. We want that /x - 2/ < 1 and also that /x - 2/ < e/65. So we simply pick d = min{1, e/65} > 0, then we should be done.

DeepSea
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Given $\varepsilon>0$ we have to find a $\delta>0$ such that $|x^4-16|< \varepsilon$ whenever $0<|x-2|<\delta$. Let $0<|x-2|<\min(1,\varepsilon/500)$ ($65$ works but just for twist the argument), so $|x|<3$ (why?) and then \begin{align}|x^4-16|= |x-2||x^3+2x^2+4x+8|\le|x-2|(|x|^3+2|x|^2+4|x|+8)\\ \le|x-2|(65)\\ <\varepsilon\end{align}

Jose Antonio
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