Error in computing a limit I was given the following limit to compute,
$$\lim_{k\to\infty}\left[k-\sqrt{k^{2}+1}\right]$$
My approach:
$$= \lim_{k\to\infty}\left[k-\sqrt{k^{2}(1+k^{-2})}\right]$$
$$= \lim_{k\to\infty}[k-k]=0$$
So the following evaluates to $0$. But my book gives that answer is $\frac{-1}{2}$. Where did my process go wrong?

Another standard approach is to use the Binomial expansion. For large positive $k,$
$$k-\sqrt{k^{2}+1}= k-\sqrt{k^{2}\left(1+k^{-2}\right)} = k-k\left(1+k^{-2}\right)^{1/2}$$
$$ = k - k\left[1+ \left( \frac{1}{2}\right) \frac{1}{k^2} + \frac{\left( \frac{1}{2}\right)\left( - \frac{1}{2}\right)}{2!} \left(\frac{1}{k^2}\right)^2 + \frac{\left( \frac{1}{2}\right)\left( - \frac{1}{2}\right)\left( - \frac{3}{2}\right)}{3!} \left(\frac{1}{k^2}\right)^3 + \ldots \right]\quad (1)$$
With a little bit of manipulation one sees, for example, that
$$ -\frac{1}{k} \left(\frac{1}{1-k^2} \right) = -\frac{1}{k} \sum_{n=0}^{\infty} \left(\frac{1}{k^2}\right)^n < (1) < 0$$
and by letting $k\to\infty$, The Sandwich Theorem gives us that expression $(1)\to 0.$
