Question

Calculate: ㅤ id="TexFormula1" title="\lim_{x \rightarrow +\infty}x(\sqrt{x^{2}-1}-x)" alt="\lim_{x \rightarrow +\infty}x(\sqrt{x^{2}-1}-x)" align="absmiddle" class="latex-formula">

Answer 1

Answer:

$\displaystyle \large \boxed{ \lim_{x \rightarrow +\infty} {x\left(\sqrt{x^2-1}-x\right)}=-\dfrac{1}{2}}$

Step-by-step explanation:

Hello, please consider the following.

$\sqrt{(x^2-1)}-x\\\\=\sqrt{x^2(1-\dfrac{1}{x^2})}-x\\\\=x\left( \sqrt{1-\frac{1}{x^2}}-1\right)$

For x close to 0, we can write

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

So,

$x\left( \sqrt{1-\frac{1}{x^2}}-1\right)\\\\=x(1-\dfrac{1}{2}\dfrac{1}{x^2}+o(\dfrac{1}{x^2})-1)\\\\=-\dfrac{1}{2x}+o(\dfrac{1}{x})$

It means that

$\displaystyle \lim_{x \rightarrow +\infty} {x\left(\sqrt{x^2-1}-x\right)}\\\\=\lim_{x \rightarrow +\infty} {-\dfrac{x}{2x}}=-\dfrac{1}{2}$

Thank you