Question

Mostre que, para todo a real

$\lim\limits_{x\to a}$ (arctg x − arctg a)/(x − a) = 1/(a² + 1)

Favor não usar as regras de L'Hospital, nem conceito pronto de derivada.

Answer 1

Queremos calcular o seguinte limite:

$L=\lim\limits_{x\to a}\dfrac{\arctan x-\arctan a}{x-a}$

Para isso, vamos fazer a substituição a seguir:

$y=\arctan x\Longrightarrow x=\tan y$

De forma conveniente, vamos fazer uma troca similar para a constante $a:$

$b=\arctan a\Longrightarrow a=\tan b$

Note que, quando $x\to a$, temos $y\to b$. Então:

$L=\lim\limits_{x\to a}\dfrac{\arctan x-\arctan a}{x-a}\\\\ L=\lim\limits_{y\to b}\dfrac{y-b}{\tan y-\tan b}$

Das relações trigonométricas, sabemos que:

$\tan(\alpha-\beta)=\dfrac{\tan(\alpha)-\tan(\beta)}{1+\tan(\alpha)\tan(\beta)}\\\\ \Longrightarrow \tan(\alpha)-\tan(\beta)=\tan(\alpha-\beta)\cdot(1+\tan(\alpha)\tan(\beta))$

Usando a identidade acima no limite:

$L=\lim\limits_{y\to b}\dfrac{y-b}{\tan y-\tan b}\\\\ L=\lim\limits_{y\to b}\dfrac{y-b}{\tan(y-b)\cdot(1+\tan y\cdot\tan b)}\\\\ L=\lim\limits_{y\to b}\dfrac{y-b}{\tan(y-b)}\cdot\dfrac{1}{1+\tan y\cdot\tan b}\\\\ L=\lim\limits_{y\to b}\dfrac{y-b}{~~\dfrac{\sin(y-b)}{\cos(y-b)}~~}\cdot\dfrac{1}{1+\tan y\cdot\tan b}\\\\ L=\lim\limits_{y\to b}\dfrac{y-b}{\sin(y-b)}\cdot\dfrac{\cos(y-b)}{1+\tan y\cdot\tan b}\\\\ L=\lim_{y\to b}\left(\dfrac{\sin(y-b)}{y-b}\right)^{-1}\cdot\dfrac{\cos(y-b)}{1+\tan y\cdot\tan b}$

Podemos utilizar acima o limite fundamental: $\lim\limits_{u\to 0}\dfrac{\sin u}{u}=1$, com $u=y-b$. Assim:

$L=\lim\limits_{y\to b}\left(\dfrac{\sin(y-b)}{y-b}\right)^{-1}\cdot\dfrac{\cos(y-b)}{1+\tan y\cdot\tan b}\\\\ L=\lim\limits_{y\to b}\left(\dfrac{\sin(y-b)}{y-b}\right)^{-1}\cdot\lim\limits_{y\to b}\dfrac{\cos(y-b)}{1+\tan y\cdot\tan b}\\\\ L=\underbrace{\lim\limits_{u\to 0}\left(\dfrac{\sin u}{u}\right)^{-1}}_{1}\cdot\lim\limits_{y\to b}\dfrac{\cos(y-b)}{1+\tan y\cdot\tan b}\\\\ L=1\cdot\dfrac{\cos(b-b)}{1+\tan b\cdot\tan b}\\\\ L=1\cdot\dfrac{\cos(0)}{1+\tan^2b}\\\\ L=\dfrac{1}{1+\tan^2b}$

Além disso, sabemos que: $\tan^2\theta+1=\sec^2\theta:$

$L=\dfrac{1}{\sec^2b}=\dfrac{1}{~~\dfrac{1}{\cos^2b}~~}\\\\ L=\cos^2b$

Em termos da constante $a$ dada no enunciado:

$L=\cos^2(\arctan a)$

Tal expressão pode ser escrita como:

$L=[\cos(\arctan a)]^2=\left[\dfrac{1}{\sqrt{a^2+1}}\right]^2\\\\ L=\dfrac{1}{a^2+1}$

Logo:

$\boxed{\lim\limits_{x\to a}\dfrac{\arctan x-\arctan a}{x-a}=\dfrac{1}{a^2+1}}~~~\blacksquare$