Trouver un équivalent simple pour chaque fonction suivante en $+\infty$:
\colonnes{\solution}{3}{1}
- \question{$\frac{1}{x-1}-\frac{1}{x+1}$} \reponse{$\frac{1}{x-1}-\frac{1}{x+1}=\frac{2}{x^{2}-1} \underset{x \rightarrow+\infty}{ } \frac{2}{x^{2}}$, car $x^{2}-1 \underset{x \rightarrow+\infty}{ } x^{2} .$}
- \question{$x+1+\ln(x)$} \reponse{$x+1+\ln (x) \underset{x \rightarrow+\infty}{-} x$, car $\lim\limits_{x \rightarrow+\infty} \frac{x+1+\ln (x)}{x}=\lim\limits_{x \rightarrow+\infty} \frac{x \times\left(1+\frac{1}{x}+\frac{\ln (x)}{x}\right)}{x}=\lim\limits_{x \rightarrow+\infty} 1+\frac{1}{x}+\frac{\ln (x)}{x}=1 .$}
- \question{$\sin\left(\frac{1}{\sqrt{x+1}}\right)$} \reponse{ $\sin (u) \underset{u \rightarrow 0}{\sim} u \Rightarrow \sin \left(\frac{1}{\sqrt{x+1}}\right) \underset{x \rightarrow+\infty}{\sim} \frac{1}{\sqrt{x+1}} \underset{x \rightarrow+\infty}{\sim} \frac{1}{\sqrt{x}}$}
- \question{$\sqrt{x+1}-\sqrt{x-1}$}
\reponse{ On a :
$\sqrt{x+1}-\sqrt{x-1}=\frac{(\sqrt{x+1}-\sqrt{x-1})(\sqrt{x+1}+\sqrt{x-1})}{\sqrt{x+1}+\sqrt{x-1}}=\frac{(x+1)-(x-1)}{\sqrt{x+1}+\sqrt{x-1}}=\frac{2}{\sqrt{x+1}+\sqrt{x-1}}$
et $\sqrt{x+1}+\sqrt{x-1} \underset{x \rightarrow+\infty}{\sim } 2 \sqrt{x}$ car :
$\lim\limits_{x \rightarrow+\infty} \frac{\sqrt{x+1}+\sqrt{x-1}}{2 \sqrt{x}}=\lim\limits_{x \rightarrow+\infty} \frac{1}{2} \sqrt{\frac{x+1}{x}}+\frac{1}{2} \sqrt{\frac{x-1}{x}}=\lim\limits_{x \rightarrow+\infty} \frac{1}{2}+\frac{1}{2}=1$ Aussi : $$ \sqrt{x+1}-\sqrt{x-1} \underset{x \rightarrow+\infty}{\sim } \frac{2}{2 \sqrt{x}}=\frac{1}{\sqrt{x}} $$} - \question{$\ln\left(\frac{x^2+x+2}{x^2+x-1}\right)$ }
\reponse{$\frac{x^{2}+x+2}{x^{2}+x-1}=\frac{x^{2} \times\left(1+\frac{1}{x}+\frac{2}{x^{2}}\right)}{x^{2} \times\left(1+\frac{1}{x}-\frac{1}{x^{2}}\right)}=\frac{1+\frac{1}{x}+\frac{2}{x^{2}}}{1+\frac{1}{x}-\frac{1}{x^{2}}}$}
$$
\frac{1+u+2 u^{2}}{1+u-u^{2}}=1+3 u^{2}+u^{2} . \varepsilon(u)
$$
$\ln (1+v)=v-\frac{v^{2}}{2}+v^{2} . \underset{v \neq 0}{\varepsilon(v)} \Rightarrow \ln \left(\frac{1+u+2 u^{2}}{1+u-u^{2}}\right)=3 u^{2}+u^{2} . \varepsilon(u)$
Ainsi : $\ln \left(\frac{x^{2}+x+2}{x^{2}+x-1}\right)=\frac{3}{x^{2}}+\frac{1}{x^{2}} \cdot \varepsilon\left(\frac{1}{x}\right)$ ou encore $\ln \left(\frac{x^{2}+x+2}{x^{2}+x-1}\right) \underset{x \rightarrow+\infty}{\sim} \frac{3}{x^{2}}$
1
\(\frac{1}{x-1}-\frac{1}{x+1}\)
2
\(x+1+\ln(x)\)
3
\(\sin\left(\frac{1}{\sqrt{x+1}}\right)\)
4
\(\sqrt{x+1}-\sqrt{x-1}\)
5
\(\ln\left(\frac{x^2+x+2}{x^2+x-1}\right)\)