Evaluating $\lim_{x\to\infty}(\sqrt{x^2+1} - x)(x+1) $

Question

The question is $$\lim_{x\to\infty}(\sqrt{x^2+1} - x)(x+1) $$ I know the answer is $\frac{1}{2}$ and I found it using this equality :

$$(\sqrt{x^2+1} - x)(x+1) = \frac{x+1}{\sqrt{x^2+1} + x}$$

But is there any other way to solve this? Any hints would be appreciated.

Answer 1

Just for fun, try letting $x=\tan\theta=\sin\theta/\cos\theta$ with $\theta\to\pi/2^-$, and use

$$\sqrt{\tan^2\theta+1}=\sqrt{\sec^2\theta}=\sec\theta={1\over\cos\theta}$$

so that

$$\begin{align} (\sqrt{x^2+1}-x)(x+1) &=\left({1\over\cos\theta}-{\sin\theta\over\cos\theta}\right)\left({\sin\theta\over\cos\theta}+1\right)\\ &={(1-\sin\theta)(\sin\theta+\cos\theta)\over\cos^2\theta}\\ &={(1-\sin\theta)(\sin\theta+\cos\theta)\over1-\sin^2\theta}\\ &={\sin\theta+\cos\theta\over1+\sin\theta} \end{align}$$

We get

$$\lim_{x\to\infty}(\sqrt{x^2+1}-x)(x+1)=\lim_{\theta\to\pi/2^-}{\sin\theta+\cos\theta\over1+\sin\theta}={1+0\over1+1}={1\over2}$$

Answer 2

Use binomial series:

$\sqrt{1+\dfrac1{x^2}}=1+\dfrac1{2x^2}\cdots$

Addendum with further explanation:

$\lim\limits_{x\to\infty}(\sqrt{x^2+1} - x)(x+1) $

$=\lim\limits_{x\to\infty}\left(\sqrt{1+\dfrac1{x^2}}-1\right)x(x+1)$

$=\lim\limits_{x\to\infty}\left(1+\dfrac1{2x^2}\cdots-1\right)(x^2+x)$

$=\lim\limits_{x\to\infty}\left(\dfrac1{2x^2}\cdots\right)(x^2+x).$

Can you take it from here?