Question

Calculati :
$\lim_{x \to \infty} x^{4} *( e^{ \frac{1}{ x^{2} +1} }- e^{ \frac{1}{ x^{2} } } } } )$

Answer 1

Scriem $x^4\left(e^{\frac{1}{x^2+1}}-e^{\frac{1}{x^2}}\right)=e^{\frac{1}{x^2}}\cdot\frac{e^{\frac{1}{x^2+1}-\frac{1}{x^2}}-1}{\frac{1}{x^4}}=e^{\frac{1}{x^2}}\cdot\frac{e^{\frac{-1}{x^4+x^2}}-1}{\frac{1}{x^4}}$

Apoi, $e^\frac{1}{x^2}\rightarrow e^0=1,$ iar $\frac{e^{\frac{-1}{x^4+x^2}}-1}{\frac{1}{x^4}}=\frac{e^{\frac{-1}{x^4+x^2}}-1}{\frac{-1}{x^4+x^2}}\cdot\frac{-x^4}{x^4+x^2}\rightarrow1\cdot (-1)=-1$.

Am folosit limita fundamentala $\lim_{x\to 0}\frac{e^x-1}{x}=1.$