Olá, Dario.
[tex]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[/tex]
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[tex]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)
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\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}:[/tex]
[tex]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[/tex]
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[tex]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[/tex]
