# Solving limit with two unknown constants [closed]

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.

• what have you tried? Jan 24 at 20:02
• I was thinking I had to write is as a fraction where the denominator goes to infinity and numerator doesn't, so I tried multuplying, diving with the conjugate but the numerator still had x with a higher power. Jan 24 at 20:07
• Replace $x+1/2$ by $y$. So you have $\lim_{y\to\infty} \sqrt{y^2-1/4}-A(y-1/2)-B =0$. Jan 24 at 20:15
• No, thanks for the help Darshan Jan 24 at 20:37
• @Olp112 No, Asked just I wonder if I missed something..... Jan 24 at 20:39

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 $$B$$.

• May I know how can I find the value of $B$ after substituting for the value of $A = 1$? Jan 24 at 20:42
• @DarshanP. We can use $(a-b)(a+b)=a^2-b^2$ to get $B=\lim_{x\to\infty} \sqrt{x^2+x}-x = \lim_{x\to\infty} \frac{x}{\sqrt{x^2+x}+x}=\lim_{x\to\infty} \frac{1}{\sqrt{1+\frac{1}{x}}+1}=\frac{1}{2}$.
– Snaw
Jan 24 at 20:47
• Yes, that's true, Nice! $\to + 1$ Jan 24 at 20:51

\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.
• Though the solution may look good, I still wonder if the limit will always be a negative zero (I mean $0^-$) Jan 24 at 20:35
• Why is B=1/2? x-Ax-B=0, I understand why A=1 but why B=1/2? Jan 24 at 20:46
• You can right it as as $$(x + \frac 12 ........\text{terms tending zero}) - Ax - B = (x - Ax) + (\frac 12 - B) + \text{terms tending to zero}$$ now in order to have the answer of limit as zero we must have the value of $A = 1$ and $B = \frac 12$ and rest of terms will tend to zero so as in net our answer of limit will also be zero. Jan 24 at 20:49

$$((x+1/2)^2-1/4)^{1/2}-Ax-B;$$

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 $$B=1/2.$$