Question

Questões de cálculo I envolvendo o teorema do confronto, segue em anexo as questões.

PS: Quero respostas concretas e bem explicadas (o passo a passo mesmo).

Answer 1

Olá, Dario.

$35.\ \begin{cases}f(x) \leq M \\-f(x)\leq M \Rightarrow f(x) >= -M \end{cases}\\\\\\ \Rightarrow -M \leq f(x) \leq M \Rightarrow \\\\ \lim\limits_{x\to0}-Mx^2 \leq \lim\limits_{x\to0}x^2f(x) \leq \lim\limits_{x\to0}Mx^2 \Rightarrow \\\\ 0 \leq \lim\limits_{x\to0}x^2f(x) \leq 0 \\\\ \text{Pelo teorema do "sandu\'iche":}\\\\ \lim\limits_{x\to0}x^2f(x)=0$
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$36.\ |f(x)| \leq M \Rightarrow -M \leq f(x) \leq M \Rightarrow \\\\ -Mg(x) \leq f(x)g(x) \leq Mg(x) \Rightarrow\\\\ \lim\limits_{x\to a}-Mg(x) \leq \lim\limits_{x\to a}f(x)g(x) \leq \lim\limits_{x\to a}Mg(x) \Rightarrow\\\\ M\lim\limits_{x\to a}-g(x) \leq \lim\limits_{x\to a}f(x)g(x) \leq M\lim\limits_{x\to a}g(x) \\\\ \text{Como }\lim\limits_{x\to a}|g(x)|=0 \Rightarrow \begin{cases} \lim\limits_{x\to a}g(x)=0 \\ \lim\limits_{x\to a}-g(x)=0 \end{cases}:$

$M\cdot0 \leq \lim\limits_{x\to a}f(x)g(x) \leq M\cdot0 \Rightarrow\\\\ 0 \leq \lim\limits_{x\to a}f(x)g(x) \leq 0\\\\ \text{Pelo teorema do "sandu\'iche":}\\\\ \lim\limits_{x\to a}f(x)g(x)=0$
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$37.\ |f(x)| \leq k|x-a| \Rightarrow\\\\ -k|x-a| \leq f(x) \leq k|x-a| \Rightarrow \\\\ \lim\limits_{x\to a}-k|x-a| \leq \lim\limits_{x\to a}f(x) \leq \lim\limits_{x\to a}k|x-a| \Rightarrow \\\\ 0 \leq \lim\limits_{x\to a}f(x) \leq 0\\\\ \text{Pelo teorema do "sandu\'iche":}\\\\ \lim\limits_{x\to a}f(x)=0$