Question

Calculate: ㅤ
`\lim_(x \rightarrow +\infty)x(\sqrt{x^(2)-1}-x)`

Answer 1

`\displaystyle \large \boxed{ \lim_(x \rightarrow +\infty) {x\left(√(x^2-1)-x\right)}=-(1)/(2)}`

Explanation:

Hello, please consider the following.

`√((x^2-1))-x\\\\=\sqrt{x^2(1-(1)/(x^2))}-x\\\\=x\left( \sqrt{1-(1)/(x^2)}-1\right)`

For x close to 0, we can write

`√(1+x)=1+(1)/(2)x-(1)/(8)x^2+o(x^2)\\\\\ \text{x tends to } +\infty \text{ means }(1)/(x) \text{ tends to 0}\\\\\text{So, when }(1)/(x)\text{ is close to 0, we can write.}\\\\\sqrt{1-(1)/(x^2)}=1-(1)/(2)(1)/(x^2)-(1)/(8)(1)/(x^4)+o((1)/(x^4))`

So,

`x\left( \sqrt{1-(1)/(x^2)}-1\right)\\\\=x(1-(1)/(2)(1)/(x^2)+o((1)/(x^2))-1)\\\\=-(1)/(2x)+o((1)/(x))`

It means that

`\displaystyle \lim_(x \rightarrow +\infty) {x\left(√(x^2-1)-x\right)}\\\\=\lim_(x \rightarrow +\infty) {-(x)/(2x)}=-(1)/(2)`

Thank you