2. prove using the $\\epsilon - \\delta$ definition, that\n(a) $\\lim_{x\\to - 4}\\frac{16 - x^{2}}{4 +…

Explanation:

Step1: Simplify the expression

For part (a), we first simplify (\frac{16 - x^{2}}{4 + x}). Using the difference - of - squares formula (a^{2}-b^{2}=(a + b)(a - b)), where (a = 4) and (b=x), we have (\frac{16 - x^{2}}{4 + x}=\frac{(4 + x)(4 - x)}{4 + x}=4 - x) for (x\neq - 4).

We want to show that for every (\epsilon>0), there exists a (\delta>0) such that if (0<|x-(-4)|=|x + 4|<\delta), then (|(4 - x)-8|=| - x - 4|=|x + 4|<\epsilon).

Let (\delta=\epsilon).

Step2: Verify the (\epsilon-\delta) condition

Suppose (0<|x + 4|<\delta). Then (| \frac{16 - x^{2}}{4 + x}-8|=|4 - x-8|=|-(x + 4)|=|x + 4|).

Since (|x + 4|<\delta) and (\delta=\epsilon), we have (| \frac{16 - x^{2}}{4 + x}-8|<\epsilon)

For part (b), we use polynomial long - division or the fact that if (f(x)=2x^{4}-6x^{3}+x^{2}+3), then (f(x)=(x - 1)q(x)+r). By the Remainder Theorem, (r = f(1)).

(f(1)=2\times1^{4}-6\times1^{3}+1^{2}+3=2-6 + 1+3=0). So (2x^{4}-6x^{3}+x^{2}+3=(x - 1)(2x^{3}-4x^{2}-3x - 3))

We want to show that for every (\epsilon>0), there exists a (\delta>0) such that if (0<|x - 1|<\delta), then (|\frac{2x^{4}-6x^{3}+x^{2}+3}{x - 1}-(-8)|=|2x^{3}-4x^{2}-3x - 3 + 8|=|2x^{3}-4x^{2}-3x + 5|)

First, assume (|x - 1|<1), so (0<x<2)

(2x^{3}-4x^{2}-3x + 5=2x^{2}(x - 2)-3x + 5)

(\left|2x^{3}-4x^{2}-3x + 5\right|=\left|2x^{2}(x - 2)-3x + 5\right|)

If (|x - 1|<\delta), we rewrite (2x^{3}-4x^{2}-3x + 5) as (2(x - 1 + 1)^{3}-4(x - 1+1)^{2}-3(x - 1 + 1)+5)

Expanding:

[ \begin{align*} 2(x - 1 + 1)^{3}&=2\left[(x - 1)^{3}+3(x - 1)^{2}+3(x - 1)+1\right]\ &=2(x - 1)^{3}+6(x - 1)^{2}+6(x - 1)+2\ -4(x - 1 + 1)^{2}&=-4\left[(x - 1)^{2}+2(x - 1)+1\right]\ &=-4(x - 1)^{2}-8(x - 1)-4\ -3(x - 1 + 1)+5&=-3(x - 1)-3 + 5=-3(x - 1)+2 \end{align*} ]

Combining terms:

[ \begin{align*} &2(x - 1)^{3}+6(x - 1)^{2}+6(x - 1)+2-4(x - 1)^{2}-8(x - 1)-4-3(x - 1)+2\ =&2(x - 1)^{3}+2(x - 1)^{2}-5(x - 1) \end{align*} ]

(\left|\frac{2x^{4}-6x^{3}+x^{2}+3}{x - 1}+8\right|=\left|2x^{3}-4x^{2}-3x + 5\right|=\left|2(x - 1)^{3}+2(x - 1)^{2}-5(x - 1)\right|)

(\leq2|x - 1|^{3}+2|x - 1|^{2}+5|x - 1|)

Let (\delta=\min\left{1,\frac{\epsilon}{9}\right})

If (|x - 1|<\delta), then (|x - 1|<1) and (|x - 1|<\frac{\epsilon}{9})

(2|x - 1|^{3}+2|x - 1|^{2}+5|x - 1|\leq2|x - 1|+2|x - 1|+5|x - 1|=9|x - 1|)

Since (|x - 1|<\delta) and (\delta=\frac{\epsilon}{9}), we have (9|x - 1|<\epsilon)

Answer:

(a) For the limit (\lim_{x\rightarrow - 4}\frac{16 - x^{2}}{4 + x}), by setting (\delta = \epsilon), the (\epsilon-\delta) definition of the limit is satisfied.

(b) For the limit (\lim_{x\rightarrow1}\frac{2x^{4}-6x^{3}+x^{2}+3}{x - 1}), by choosing (\delta=\min\left{1,\frac{\epsilon}{9}\right}), the (\epsilon-\delta) definition of the limit is satisfied.