Question

8.Determinar os seguintes limites com auxílio das regras de L'Hospital:

lim x² (e¹/ˣ -1)
x⇒+∞

Answer 1

$^{~lim}_{x\to\infty}~~x^2\left(e^{\frac{1}{x}}-1\right)~=\\\\\\Temos~uma~indeterminacao~\infty\,.~0,~logo~ainda~nao~podemos~aplicar\\l'Hospital.\\\\\\Vamos~reorganizar~a~expressao~para~que~apareca~\frac{0}{0}\\\\\\=~^{~lim}_{x\to\infty}~~\frac{\left(e^{\frac{1}{x}}-1\right)}{\frac{1}{x^2}}\\\\\\Agora~sim,~a~indeterminacao~\acute{e}~do~tipo~\frac{0}{0}\\\\Aplicando~l'Hospital:$

$^{~lim}_{x\to\infty}~~\frac{\left(e^{\frac{1}{x}}-1\right)}{\frac{1}{x^2}}~=~^{~lim}_{x\to\infty}~~\frac{\left(e^{\frac{1}{x}}-1\right)'}{\left(\frac{1}{x^2}\right)'}\\\\\\^{~lim}_{x\to\infty}~~\frac{\left(e^{\frac{1}{x}}-1\right)}{\frac{1}{x^2}}~=~^{~lim}_{x\to\infty}~~\frac{-\frac{1}{x^2}~.~e^{\frac{1}{x}}}{\frac{-2x}{x^4}}\\\\\\^{~lim}_{x\to\infty}~~\frac{-\frac{1}{x^2}~.~e^{\frac{1}{x}}}{\frac{-2x}{x^4}}~=~^{~lim}_{x\to\infty}~~\frac{-e^{\frac{1}{x}}~.~x^4}{-x^2~.~2x}$

$^{~lim}_{x\to\infty}~~\frac{-e^{\frac{1}{x}}~.~x^4}{-x^2~.~2x}~=~^{~lim}_{x\to\infty}~~\frac{e^{\frac{1}{x}}~.~x}{2}\\\\\\^{~lim}_{x\to\infty}~~\frac{e^{\frac{1}{x}}~.~x}{2}~=~\frac{1~.~\infty}{2}~=~\boxed{\infty}$