Matemática III

51.
$\\[0\parindent]\lim_{\alpha \to \infty} \frac{\ln(1+e^\alpha)}{\alpha}=\\[2\baselineskip]1: = \lim_ {\alpha \to -\infty}\frac{\ln(1+e^\alpha)}{\alpha} = 0 \\[2\baselineskip] 2: \lim_ {\alpha \to +\infty} \frac{\ln(e^\alpha)}{\alpha} \le \\[2\baselineskip] \le \lim_ {\alpha \to +\infty} \frac{\ln(1+e^\alpha)}{\alpha} \le \\[2\baselineskip] \le \lim_ {\alpha \to +\infty} \frac{\ln(e^{\alpha+1})}{\alpha} \\[2\baselineskip] \lim_ {\alpha \to +\infty} \frac{\alpha}{\alpha}=1 \le \\[2\baselineskip] \le \lim_ {\alpha \to +\infty} \frac{\ln(1+e^\alpha)}{\alpha} \le \\[2\baselineskip] \le \lim_ {\alpha \to +\infty} \frac{\alpha+1}{\alpha}=1$

52.
$\\[0\parindent]\lim_{x \to 0} \frac{\sin \alpha x}{\sin \beta x}=\\[2\baselineskip] = \lim_{x \to 0} \frac{\sin \alpha x}{\alpha x}\frac{\beta x}{\sin \beta x}\frac{\alpha x}{\beta x}=\\[2\baselineskip] = \frac{\alpha}{\beta}$

53.
$\\[0\parindent]\lim_{x \to \infty}\frac{\alpha^x-1}{x}, (\alpha>1)=\\[2\baselineskip] 1: =\lim_{x \to +\infty}\frac{(1+b)^x-1}{x}, (0<b<1)\\[2\baselineskip] \lim_{n \to +\infty}\frac{(1+b)^n-1}{n}\le\lim_{x \to +\infty}\frac{(1+b)^x-1}{x}\le\lim_{n \to +\infty}\frac{(1+b)^{n+1}-1}{n+1}, n=\lfloor x\rfloor \\[2\baselineskip] \lim_{n \to +\infty}\frac{(1+b)^n-1}{n} = \lim_{n \to +\infty}\frac{\sum_{i=1}^{n}\binom{n}{i}1^{n-i}b^i-1}{n} // Newton binomy\\[2\baselineskip]$
$max(\binom{n}{i}) = \binom{n}{\lfloor \frac{n}{2}\rfloor }, i= 1..n\\[2\baselineskip] \binom{n}{\lfloor \frac{n}{2}\rfloor } = \frac{n!}{\lceil \frac{n}{2} \rceil ! \lfloor \frac{n}{2} \rfloor !}\\[2\baselineskip] \lim_ {n \to +\infty} \frac{n!}{\lceil \frac{n}{2} \rceil ! \lfloor \frac{n}{2} \rfloor !} . \frac{1}{p^{\lfloor\frac{n}{2}\rfloor}}= , \frac{1}{p}=b\\[2\baselineskip] = \frac{n}{\lceil \frac{n-1}{2} \rceil p^2} . \frac{n-1}{\lceil \frac{n-3}{2} \rceil p} . \frac{n-2}{\lceil \frac{n-5}{2} \rceil p} ... \frac{\lceil \frac{n+3}{2} \rceil}{2 p} . \frac{\lceil \frac{n+2}{2} \rceil}{p} = \\[2\baselineskip] =+\infty$

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Matemática III

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