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.


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


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



Use binomial series:


Addendum with further explanation:

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




Can you take it from here?

  • $\begingroup$ I did this as my first attempt but it gave me $ 0 \times \infty $ and I didn't know what to do next. $\endgroup$ – Arsalan Hasanvand Jan 17 at 0:15
  • $\begingroup$ guess I should read some book to find out how to resolve the indeterminacy of this kind, thank you btw. $\endgroup$ – Arsalan Hasanvand Jan 17 at 0:20
  • $\begingroup$ You're welcome; I added further explanation $\endgroup$ – J. W. Tanner Jan 17 at 1:32
  • $\begingroup$ yeah got it! tnx $\endgroup$ – Arsalan Hasanvand Jan 17 at 4:26

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