Why does the limit of this function have such a different result than what I expected?

Question

So I found this function$$f(x)=x\sqrt{x^2-1}$$Why does it appear that as $x$ gets greater, the function approaches $x^2-0.5$ $$\lim_{x\to\infty}f(x)=x^2-\frac12$$I don't understand where the $(-0.5)$ comes from, the only logical conclusion I could make is that it has to do with the $(-1)$ under the square root but I would've assumed it would become negligible eventually, with the function tending towards $x^2$. Why does the function behave like that?

Answer 1

From the binomial theorem, for $x>0$ $$\sqrt{x^2-1}=x(1-1/x^2)^{1/2}=x\left(1-\frac{1}{2x^2}+\textrm{higher terms in $1/x^2$}\right)$$ and so $$f(x)=x^2-\frac12+\textrm{higher terms in $1/x^2$}$$ and these higher terms approach zero as $x\to\infty$.

Answer 2

see that $$\sqrt{x^2-1}=x\sqrt{1-\frac{1}{x^2}}= x\left(1-\frac{1}{2x^2}+\frac{1}{6x^4} \cdots \right)$$

Answer 3

$$f(x)=x\sqrt{x^2-1}=x^2\sqrt{1-\frac 1{x^2}}$$ Now for small$\epsilon$, $\sqrt{1-\epsilon}\sim 1-\frac \epsilon 2$. Then $$f(x)=x\sqrt{x^2-1}\sim x^2\left(1-\frac 1 {2x^2} \right)=x^2-\frac 12$$

Answer 4

It is cleaner to write

$$\lim_{x\to\infty}\left(x\sqrt{x^2-1}-x^2\right)=-\frac12,$$

which is easily established by

$$\lim_{x\to\infty}\left(x\sqrt{x^2-1}-x^2\right)=\lim_{x\to\infty}\frac{(x\sqrt{x^2-1})^2-(x^2)^2}{x\sqrt{x^2-1}+x^2}=-\lim_{x\to\infty}\frac 1{\sqrt{1-\dfrac1{x^2}}+1}.$$