Question

Preciso de ajuda para resolver esse lim
$lim \: \frac{ \sin(x) - \sin(p) }{x \: - p} \\ x - > p$
​

Answer 1

Olá!

$\displaystyle\lim_{x\to{p}}\dfrac{\sin(x)-\sin(p)}{x-p}$

Façamos a seguinte substituição de variável:

$x-p=h\longrightarrow{x}-h=p$

Por conseguinte, temos que: $x\to{p}\Rightarrow{h}\to0$.

Segue-se:

$\displaystyle\lim_{x\to{p}}\dfrac{\sin(x)-\sin(p)}{x-p}=\displaystyle\lim_{h\to0}\dfrac{\sin(x)-\sin(x-h)}{h}$

Pela relação do seno da soma de arcos:

$\sin(x-h)=\sin(x).\cos(h)-\sin(h).\cos(x)$

Então:

$\displaystyle\lim_{x\to{p}}\dfrac{\sin(x)-\sin(p)}{x-p}=\displaystyle\lim_{h\to0}\dfrac{\sin(x)-(\sin(x).\cos(h)-\sin(h).\cos(x))}{h}$

$\displaystyle\lim_{x\to{p}}\dfrac{\sin(x)-\sin(p)}{x-p}=\displaystyle\lim_{h\to0}\dfrac{\sin(x)-\sin(x).\cos(h)+\sin(h).\cos(x)}{h}$

$\displaystyle\lim_{x\to{p}}\dfrac{\sin(x)-\sin(p)}{x-p}=\displaystyle\lim_{h\to0}\dfrac{\sin(x)(1-\cos(h))+\sin(h).\cos(x)}{h}$

$\displaystyle\lim_{x\to{p}}\dfrac{\sin(x)-\sin(p)}{x-p}=\displaystyle\lim_{h\to0}\dfrac{\sin(x)(1-\cos(h))}{h}+\displaystyle\lim_{h\to0}\dfrac{\sin(h).\cos(x)}{h}$

As funções $\sin(x)$ e $\cos(x)$ independem de $h$.

$\displaystyle\lim_{x\to{p}}\dfrac{\sin(x)-\sin(p)}{x-p}=\sin(x)\displaystyle\lim_{h\to0}\dfrac{1-\cos(h)}{h}+\cos(x)\displaystyle\lim_{h\to0}\dfrac{\sin(h)}{h}$

$\displaystyle\lim_{x\to{p}}\dfrac{\sin(x)-\sin(p)}{x-p}=\sin(x)\displaystyle\lim_{h\to0}\left(\dfrac{1-\cos(h)}{h}\!\cdot\!\dfrac{1+\cos(h)}{1+\cos(h)}\right)\!+\cos(x)\!\cdot\!1$

$\displaystyle\lim_{x\to{p}}\dfrac{\sin(x)-\sin(p)}{x-p}=\sin(x)\displaystyle\lim_{h\to0}\dfrac{1-\cos^2(h)}{h(1+\cos(h))}+\cos(x)$

$\displaystyle\lim_{x\to{p}}\dfrac{\sin(x)-\sin(p)}{x-p}=\sin(x)\displaystyle\lim_{h\to0}\dfrac{\sin^2(h)}{h(1+\cos(h))}+\cos(x)$

$\displaystyle\lim_{x\to{p}}\dfrac{\sin(x)-\sin(p)}{x-p}=\sin(x)\displaystyle\lim_{h\to0}\dfrac{\sin(h)\!\cdot\!\sin(h)}{h(1+\cos(h))}+\cos(x)$

$\displaystyle\lim_{x\to{p}}\dfrac{\sin(x)-\sin(p)}{x-p}=\sin(x)\!\left(\!\displaystyle\lim_{h\to0}\dfrac{\sin(h)}{h}\!\right)\!\!\left(\!\displaystyle\lim_{h\to0}\dfrac{\sin(h)}{1+\cos(h)}\!\right)\!+\cos(x)$

$\displaystyle\lim_{x\to{p}}\dfrac{\sin(x)-\sin(p)}{x-p}=\sin(x)\!\cdot\!1\!\cdot\!\dfrac{0}{1+1}+\cos(x)$

$\displaystyle\lim_{x\to{p}}\dfrac{\sin(x)-\sin(p)}{x-p}=\cos(x)$

Qualquer dúvida, comente! Bons estudos!