Question

limite
$lim \: (x ln(x + 1) - x ln(x))$
​

Answer 1

Resposta:  1.

Explicação passo-a-passo:

Calcular o limite:

    $\mathsf{\underset{x\to \infty}{\ell im}~\big[x\,\ell n(x+1)-x\,\ell n(x)\big]}$

Aplicando propriedades de logaritmos para x > 0, podemos reescrever como

      $\mathsf{=\underset{x\to \infty}{\ell im}~x\cdot \big[\ell n(x+1)-\ell n(x)\big]}\\\\\\ \mathsf{=\underset{x\to \infty}{\ell im}~x\cdot \ell n\left(\dfrac{x+1}{x}\right)}\\\\\\ \mathsf{=\underset{x\to \infty}{\ell im}~x\cdot \ell n\left(\dfrac{x}{x}+\dfrac{1}{x}\right)}\\\\\\ \mathsf{=\underset{x\to \infty}{\ell im}~x\cdot \ell n\left(1+\dfrac{1}{x}\right)}\\\\\\ \mathsf{=\underset{x\to \infty}{\ell im}~\dfrac{\ell n\left(1+\frac{1}{x}\right)}{\frac{1}{x}}}$

Faça a seguinte mudança de variável:

    $\mathsf{\dfrac{1}{x}=u,~~u>0\quad\Longrightarrow\quad x=\dfrac{1}{u}}$

e $\mathsf{u\to 0^+}$  quando $\mathsf{x\to \infty}.$

Substituindo, o limite fica

    $\mathsf{=\underset{u\to 0^+}{\ell im}~\dfrac{\ell n(1+u)}{u}}$

Ainda temos uma indeterminação do tipo "0/0". Mas agora fica muito mais simples de resolver aplicando a regra de L'Hôpital:

    $\mathsf{=\underset{u\to 0^+}{\ell im}~\dfrac{\frac{d}{du}\,\ell n(1+u)}{\frac{d}{du}\,u}}\\\\\\ \mathsf{=\underset{u\to 0^+}{\ell im}~\dfrac{~\frac{1}{1+u}~}{1}}\\\\\\ \mathsf{=\underset{u\to 0^+}{\ell im}~\dfrac{1}{1+u}}\\\\\\ \mathsf{=\dfrac{1}{1+0}}\\\\\\ \mathsf{=1\quad\longleftarrow\quad resposta.}$

Dúvidas? Comente.

Bons estudos! :-)