Ensino Superior ⇒ Cálculo 1 - Limites Tópico resolvido

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Resposta

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3

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[tex3]
g(x)=6x\\
(f\!\circ\! g)'(x)=f'(g(x))\cdot g'(x)=6f'(g(x))\\
h(x)=\sin(2x)+\sinh(2x)\implies h'(x)=2\cos(2x)+2\cosh(2x)\\[24pt]

\forall x\neq 0,\,\dfrac{f(6x)}{\sin(2x)+\sinh(2x)}=\dfrac{f\!\circ \!g(x)-f\!\circ\! g(0)}{x-0}\cdot\dfrac{x-0}{h(x)-h(0)}\\[12pt]
\lim\limits_{x\to0}\dfrac{f\!\circ \!g(x)-f\!\circ\! g(0)}{x-0}=(f\!\circ\! g)'(0)=6f'(0)=12\\[12pt]
\lim\limits_{x\to0}\dfrac{h(x)-h(0)}{x-0}=h'(0)=4\\[12pt]
\lim\limits_{x\to0}\dfrac{f(6x)}{\sin(2x)+\sinh(2x)}=12\times \dfrac{1}{4}=3

[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]
g(x)=6x\\
(f\!\circ\! g)'(x)=f'(g(x))\cdot g'(x)=6f'(g(x))\\
h(x)=\sin(2x)+\sinh(2x)\implies h'(x)=2\cos(2x)+2\cosh(2x)\\[24pt]

\forall x\neq 0,\,\dfrac{f(6x)}{\sin(2x)+\sinh(2x)}=\dfrac{f\!\circ \!g(x)-f\!\circ\! g(0)}{x-0}\cdot\dfrac{x-0}{h(x)-h(0)}\\[12pt]
\lim\limits_{x\to0}\dfrac{f\!\circ \!g(x)-f\!\circ\! g(0)}{x-0}=(f\!\circ\! g)'(0)=6f'(0)=12\\[12pt]
\lim\limits_{x\to0}\dfrac{h(x)-h(0)}{x-0}=h'(0)=4\\[12pt]
\lim\limits_{x\to0}\dfrac{f(6x)}{\sin(2x)+\sinh(2x)}=12\times \dfrac{1}{4}=3

[/tex3]