Solving limit with two unknown constants How can I find the constants $A$ and $B$ if $$\lim_{x\to\infty}\left(\sqrt{x^2+x}-Ax-B\right) = 0\;\;\;\;?$$
I'm not sure how I might get the answer for two unknowns with one equation.
 A: You're trying to find the asymptote of $\sqrt{x^2+x}$ at infinity. The standard method is to say that if
$$\lim_{x\to\infty}\sqrt{x^2+x}-Ax-B = 0$$
then surely also
$$\lim_{x\to\infty}\frac{\sqrt{x^2+x}-Ax-B}{x} = 0$$
which implies
$$A = \lim_{x\to\infty} \frac{\sqrt{x^2+x}}{x}=\lim_{x\to\infty}\sqrt{1+\frac{1}{x}}=1$$
So $A=1$ and now you can substitute this back into the original equation to find $B$.
A: $$\begin{align*}
\lim_{x\to \infty} \sqrt{x^2 + x} - Ax - B
& = \lim_{x \to \infty}x\sqrt{1 + \frac 1x} - Ax - B\\
& = \lim_{x \to \infty} x \left( 1+ \frac 12\frac 1x + ..........\text{terms with higher power of }\frac 1x\right) - Ax - B\\
& = \lim_{x \to \infty}\left(x +  \frac 12......\text{terms tending zero as x will tend to infinity}\right) - Ax - B\\
& = 0 \text{ given}\\
\implies A =1 \text{ and } B = \frac 12
\end{align*}$$

*

*How and why can I expand $\sqrt {1 + \frac 1x}$?
Because the expansion of $(1+\delta)^{\frac 12} $ is allowed if $\delta$ is really a $\delta$ (I mean small) which here $\frac 1x$ tends to be very small as $x\to\infty$. So yes! we are good to expand.

A: $((x+1/2)^2-1/4)^{1/2}-Ax-B;$
Set $y:=x+1/2,$ and consider $y \rightarrow \infty$.
$(y^2-1/4)^{1/2}-A(y-1/2)-B;$
The asymptote of $(y^2-1/4)^{1/2}$ is $y$.
Collecting corresponding terms in the asymptotic expression:
$y(1-A)+((1/2)A-B) =0$ we get
$A=1$ and $B=1/2.$
