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?

$\require{\cancel}$
Your book is incorrect. Since $(a-b)(a+b) = a^2-b^2$ we get for $k>0$
\begin{align}
k - \sqrt{k^2 +1}& = k - \sqrt{k^2 +1} \left(\frac{k + \sqrt{k^2 +1}}{k + \sqrt{k^2 +1}} \right) \\
&= \frac{\cancel{k^2} - (\cancel{k^2} +1)}{k + \sqrt{k^2 +1}} \\
&= - \frac{1}{k + k\sqrt{ 1+\frac{1}{k^2}}} \\
& = -\frac{1}{k}\cdot\frac{1}{1+\sqrt{ 1+\frac{1}{k^2}}}
\end{align}
And since both $\lim_{k \to \infty}-\frac{1}{k} =0$ and $\lim_{k \to \infty}\frac{1}{1+\sqrt{ 1+\frac{1}{k^2}}} = \frac{1}{2}$ exist, your original limit is the product of these limits. Hence
$$
\lim_{k \to \infty}k - \sqrt{k^2 +1} =0\cdot \frac12 =0
$$
