Atividade de Calculo diferencial e integral
Aplicando a regra de L'hospital, resolva:
lim de x tendendo a 0 positivo(x a terceira Inx)
fabianasantoslima:
Agora aprendi como anexar ,vou anexi-los depois vc da uma olhadinha obrigada .
Soluções para a tarefa
Respondido por
0
Olá, Fabiana,
![\lim\limits_{x\to0^{+}}x\³\ln x=\lim\limits_{x\to0^{+}}\frac{\ln x}{x^{-3}}=\\\\\\
\text{(Aplicando L'H\^opital)}\\\\\
=\lim\limits_{x\to0^{+}}\frac{(\ln x)'}{(x^{-3})'}=\lim\limits_{x\to0^{+}}\frac{x^{-1}}{-3x^{-4}}=\lim\limits_{x\to0^{+}}[-\frac13x^{-1-(-4)}]=\\\\
=-\frac13\lim\limits_{x\to0^{+}}x^{3}=\\\\
=\boxed{0}](https://tex.z-dn.net/?f=%5Clim%5Climits_%7Bx%5Cto0%5E%7B%2B%7D%7Dx%5C%C2%B3%5Cln+x%3D%5Clim%5Climits_%7Bx%5Cto0%5E%7B%2B%7D%7D%5Cfrac%7B%5Cln+x%7D%7Bx%5E%7B-3%7D%7D%3D%5C%5C%5C%5C%5C%5C%0A%5Ctext%7B%28Aplicando+L%27H%5C%5Eopital%29%7D%5C%5C%5C%5C%5C%0A%3D%5Clim%5Climits_%7Bx%5Cto0%5E%7B%2B%7D%7D%5Cfrac%7B%28%5Cln+x%29%27%7D%7B%28x%5E%7B-3%7D%29%27%7D%3D%5Clim%5Climits_%7Bx%5Cto0%5E%7B%2B%7D%7D%5Cfrac%7Bx%5E%7B-1%7D%7D%7B-3x%5E%7B-4%7D%7D%3D%5Clim%5Climits_%7Bx%5Cto0%5E%7B%2B%7D%7D%5B-%5Cfrac13x%5E%7B-1-%28-4%29%7D%5D%3D%5C%5C%5C%5C%0A%3D-%5Cfrac13%5Clim%5Climits_%7Bx%5Cto0%5E%7B%2B%7D%7Dx%5E%7B3%7D%3D%5C%5C%5C%5C%0A%3D%5Cboxed%7B0%7D "\lim\limits_{x\to0^{+}}x\³\ln x=\lim\limits_{x\to0^{+}}\frac{\ln x}{x^{-3}}=\\\\\\
\text{(Aplicando L'H\^opital)}\\\\\
=\lim\limits_{x\to0^{+}}\frac{(\ln x)'}{(x^{-3})'}=\lim\limits_{x\to0^{+}}\frac{x^{-1}}{-3x^{-4}}=\lim\limits_{x\to0^{+}}[-\frac13x^{-1-(-4)}]=\\\\
=-\frac13\lim\limits_{x\to0^{+}}x^{3}=\\\\
=\boxed{0}")
Perguntas interessantes