1.3

1. \(\lim_{x\to0}\frac{\sin(x)-x}{\arctan(x)-x}=\frac{T_{n,\sin(x),0}+E_1(x)-x}{T_{m,\arctan(x),0}+E_2(x)-x}=\lim_{x\to0}\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}+E_1\left(x\right)-x}{x-\frac{x^3}{3!}+\frac{x^5}{5!}+E_2(x)-x}\)

   We use this here \(E(x)=\frac{f^{N+1}\left(c\right)}{(N+1)!}(x-a)^{N+1}\), then \(E_1(x)=\frac{\sin^6\left(c\right)}{6!}x^6,E_2(x)=\frac{\arctan^6\left(\tilde{c}\right)}{6!}x^6\)​

   Then \(=\lim_{x\to0}\frac{x^3\left(-\frac{1}{3!}+\frac{x^2}{5!}+Ax^3\right)}{x^3\left(-\frac13+\frac{x^2}{5}+Bx^3\right)}=\frac12\)

2. Theorem (Criterion for Local Maximum and Minimum): If \(f'(a) = f''(a) = \ldots = f^{(n-1)}(a) = 0\) but \(f^{(n)}(a) \neq 0\), then:

   1. Case \(n\) is even:

      • If \(f^{(n)}(a) > 0\), \(f\) has a local minimum at \(x = a\).

      • If \(f^{(n)}(a) < 0\), \(f\) has a local maximum at \(x = a\).

        1. Case \(n\) is odd:

      • No local extremum exists.

   Proof:

   \(\lim_{x \to a} \frac{f(x) - T_{n,f,a}(x)}{(x-a)^n} = 0.\)Where \(T_{n,f,a}(x) = \sum_{i=0}^n a_i (x-a)^i\), and \(a_i = \frac{f^{(i)}(a)}{i!}\)​\(\lim_{x \to a} \frac{f(x) - f(a) - \frac{f'(a)}{1!}(x-a) - \frac{f''(a)}{2!}(x-a)^2 - \cdots - \frac{f^{(n)}(a)}{n!}(x-a)^n}{(x-a)^n} = 0\)

   We can assume \(f(a) = 0.\) If \(f(a) \neq 0\), define a new function: \(g(x) = f(x) - f(a),\)with the same conditions as \(f\).

   Then we have \(\lim_{x \to a} \frac{f(x) - \frac{f^{(n)}(a)}{n!}(x-a)^n}{(x-a)^n} = \lim_{x \to a} \left( \frac{f(x)}{(x-a)^n} - \frac{f^{(n)}(a)}{n!} \right) = 0.\)​

   Suppose \(n\) is even: \((x-a)^n > 0\) for all \(x\).

   Moreover, When \(x\) is sufficiently close to \(a\), the expressions:\(\frac{f(x)}{(x-a)^n} \text{ and } \frac{f^{(n)}(a)}{n!}\) have the same sign.

   If \(f^{(n)}(a) > 0\), then \(\frac{f^{n}\left(a\right)}{n!}>0,\frac{f(x)}{(x-a)^{n}}>0\Rightarrow f(x)>0\text{ for }x\text{ close to }a.\)

   Finally, since \(f(a)=0\), then \(f(x)>f(a)\), then \(f\) has a local minimum at \(x = a\).

3. Find the local maximum and local minimum of \(f(x) = \sin^2(x) \cos^2(x)\) in \(\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)\).

   ---

   \(f'(x) = 2\sin(x)\cos(x)\cos^2(x) + \sin^2(x) \cdot 2\cos(x)(-\sin(x))\)​

   \(f'(x) = 2\sin(x)\cos^3(x) - 2\sin^3(x)\cos(x)\)​

   \(f'(x) = 2\sin(x)\cos(x)\left(\cos^2(x) - \sin^2(x)\right)\)​

   \(f'(x) = 2\sin(x)\cos(x)\cos(2x)\)

   ---

   \(f'(x) = 0 \iff \sin(x) = 0 \quad \text{or} \quad \cos(x) = 0 \quad \text{or} \quad \cos(2x) = 0\) where \(x \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right)\)

   Since \(\cos x,\cos 2x>0\), then \(x=0\)

   ---

   \(f''(x)=2\cos^2(x)\cos(2x)-2\sin^2(x)\cos(2x)-2\sin x\cos x\cdot\sin2x\cdot2\)​

   \(f''(0) = 2 \cdot 1 > 0 \implies \text{local minimum at } x = 0\)​

4. Example: Compute \(T_{N, \sin(5x), 0}, \quad T_{N, \sin(x^3), 0}.\)

   For \(\sin(x)\)​ :

   \(\sin(x) \rightarrow T_{2N+1, \sin(x), 0}(x) = \sum_{j=0}^N \frac{(-1)^j x^{2j+1}}{(2j+1)!}.\)

   For \(\sin(5x)\)​ :

   \(\sin(5x)\rightarrow T_{2N+1,\sin(5x),0}(x)=\sum_{j=0}^{N}\frac{(-1)^{j}(5x)^{2j+1}}{(2j+1)!}=\sum_{j=0}^{N}\frac{(-1)^{j}5^{2j+1}x^{2j+1}}{(2j+1)!}\)​

   For \(\sin(x^3)\)​ :

   \(\sin(x^3)\rightarrow T_{6N+3,\sin(x^3),0}(x)=\sum_{j=0}^{N}\frac{(-1)^{j}(x^3)^{2j+1}}{(2j+1)!}=\sum_{j=0}^{N}\frac{(-1)^{j}x^{6j+3}}{(2j+1)!}\)​

   We can use this to formal prove \(\lim_{x \to 0} \frac{f(x) - g(x)}{(x-0)^n} = 0.\)

   \(f'(x) = 2\sin(x)\cos(x)\cos(2x)\)

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