8
$\begingroup$

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.

$\endgroup$
8
$\begingroup$

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

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

$\endgroup$
0
3
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.