Question

Qual o limite de
(1+h)² - 1 / h
com h --> 0?

Answer 1

$\lim\limits_{h\rightarrow0}\dfrac{(1+h)^{2}-1}{h}=\lim\limits_{h\rightarrow0}\dfrac{(1^{2}+2\cdot1\cdot h+h^{2})-1}{h}\\\\\\\lim\limits_{h\rightarrow0}\dfrac{(1+h)^{2}-1}{h}=\lim\limits_{h\rightarrow0}\dfrac{1+2h+h^{2}-1}{h}\\\\\\\lim\limits_{h\rightarrow0}\dfrac{(1+h)^{2}-1}{h}=\lim\limits_{h\rightarrow0}\dfrac{2h+h^{2}}{h}$

Colocando h em evidência:

$\lim\limits_{h\rightarrow0}\dfrac{(1+h)^{2}-1}{h}=\lim\limits_{h\rightarrow0}\dfrac{h\cdot(2+h)}{h}$

Cortando h com h:

$\lim\limits_{h\rightarrow0}\dfrac{(1+h)^{2}-1}{h}=\lim\limits_{h\rightarrow0}(2+h)$

Agora podemos substituir normalmente:

$\lim\limits_{h\rightarrow0}\dfrac{(1+h)^{2}-1}{h}=2+0\\\\\\\boxed{\boxed{\lim\limits_{h\rightarrow0}\dfrac{(1+h)^{2}-1}{h}=2}}$
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Podíamos evitar o desenvolvimento do quadrado da soma, percebendo que

$a^{2}-b^{2}=(a+b)\cdot(a-b)~~~~(produto~da~soma~pela~dif.)$

Veja:

$\lim\limits_{h\rightarrow0}\dfrac{(1+h)^{2}-1}{h}=\lim\limits_{h\rightarrow0}\dfrac{(1+h)^{2}-1^{2}}{h}\\\\\\\lim\limits_{h\rightarrow0}\dfrac{(1+h)^{2}-1}{h}=\lim\limits_{h\rightarrow0}\dfrac{(1+h+1)\cdot(1+h-1)}{h}\\\\\\\lim\limits_{h\rightarrow0}\dfrac{(1+h)^{2}-1}{h}=\lim\limits_{h\rightarrow0}\dfrac{(2+h)\cdot h}{h}\\\\\\\lim\limits_{h\rightarrow0}\dfrac{(1+h)^{2}-1}{h}=\lim\limits_{h\rightarrow0}(2+h)\\\\\\\lim\limits_{h\rightarrow0}\dfrac{(1+h)^{2}-1}{h}=2+0=2$
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Também podíamos ter percebido que esse limite é a definição da derivada de f(x) = x² em x = 1

Definição de derivada:

$f'(x)=\lim\limits_{h\rightarrow0}\dfrac{f(x+h)-f(x)}{h}$

Então, a derivada em x = 1 é dada por:

$f'(1)=\lim\limits_{h\rightarrow0}\dfrac{f(1+h)-f(1)}{h}$

Se f(x) = x², temos:

$f'(x)=2x\\\\f(1+h)=(1+h)^{2}\\\\f(1)=1^{2}=1$

Então:

$f'(1)=\lim\limits_{h\rightarrow0}\dfrac{(1+h)^{2}-1}{h}=2\cdot1=2$