Calculate $\lim_{x\to\infty}\frac{x^2}{x+1}-\sqrt{x^2+1}$ I am stuck on a limit of the indeterminate form $\infty-\infty$. I have tried many approaches, such as multiplying with conjugates etc. and I am unable to find a solution. I suspect that there is an elementary trick that I am plainly missing right here. Can anybody give me a hint or solution as to solve
$$\lim_{x\to\infty}\frac{x^2}{x+1}-\sqrt{x^2+1}$$

If
$f(x)
=\dfrac{x^2}{x+1}-\sqrt{x^2+1}
$
then
$\begin{array}\\
f(x)
&=\dfrac{x^2}{x+1}-\sqrt{x^2+1}\\
&=\dfrac{x^2+x-x}{x+1}-\sqrt{x^2+1}\\
&=x-\dfrac{x}{x+1}-\sqrt{x^2+1}\\
&=x-\dfrac{x+1-1}{x+1}-\sqrt{x^2+1}\\
&=x-1+\dfrac{1}{x+1}-\sqrt{x^2+1}\\
\text{so}\\
f(x)
&\lt x-1+\dfrac{1}{x+1}-\sqrt{x^2}\\
&= -1+\dfrac{1}{x+1}\\
\text{and}\\
f(x)
&=-1+\dfrac{1}{x+1}+x-\sqrt{x^2+1}\\
&=-1+\dfrac{1}{x+1}+(x-\sqrt{x^2+1})\dfrac{x+\sqrt{x^2+1}}{x+\sqrt{x^2+1}}\\
&=-1+\dfrac{1}{x+1}-\dfrac{1}{x+\sqrt{x^2+1}}\\
&>-1+\dfrac{1}{x+1}-\dfrac{1}{2x}\\
&=-1+\dfrac{2x-(x+1)}{2x(x+1)}\\
&=-1+\dfrac{x-1}{2x(x+1)}\\
&=-1+\dfrac{x+1-2}{2x(x+1)}\\
&=-1+\dfrac{1}{2x}-\dfrac{1}{x(x+1)}\\
\end{array}
$
so
$f(x) \to -1$.
