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

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

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

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

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

Use binomial series:

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

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

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

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

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

Can you take it from here?

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