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.
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1$\begingroup$ what have you tried? $\endgroup$– cineelJan 24 at 20:02
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$\begingroup$ 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. $\endgroup$– Olp112Jan 24 at 20:07
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3$\begingroup$ Replace $x+1/2$ by $y$. So you have $\lim_{y\to\infty} \sqrt{y^2-1/4}-A(y-1/2)-B =0 $. $\endgroup$– markvsJan 24 at 20:15
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1$\begingroup$ No, thanks for the help Darshan $\endgroup$– Olp112Jan 24 at 20:37
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$\begingroup$ @Olp112 No, Asked just I wonder if I missed something..... $\endgroup$– Darshan P.Jan 24 at 20:39
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3 Answers
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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 $B$.
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$\begingroup$ May I know how can I find the value of $B$ after substituting for the value of $A = 1$? $\endgroup$ Jan 24 at 20:42
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$\begingroup$ @DarshanP. We can use $(a-b)(a+b)=a^2-b^2$ to get $B=\lim_{x\to\infty} \sqrt{x^2+x}-x = \lim_{x\to\infty} \frac{x}{\sqrt{x^2+x}+x}=\lim_{x\to\infty} \frac{1}{\sqrt{1+\frac{1}{x}}+1}=\frac{1}{2}$. $\endgroup$– SnawJan 24 at 20:47
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$$\begin{align*} \lim_{x\to \infty} \sqrt{x^2 + x} - Ax - B & = \lim_{x \to \infty}x\sqrt{1 + \frac 1x} - Ax - B\\ & = \lim_{x \to \infty} x \left( 1+ \frac 12\frac 1x + ..........\text{terms with higher power of }\frac 1x\right) - Ax - B\\ & = \lim_{x \to \infty}\left(x + \frac 12......\text{terms tending zero as x will tend to infinity}\right) - Ax - B\\ & = 0 \text{ given}\\ \implies A =1 \text{ and } B = \frac 12 \end{align*}$$
- 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.
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$\begingroup$ Though the solution may look good, I still wonder if the limit will always be a negative zero (I mean $0^-$) $\endgroup$ Jan 24 at 20:35
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$\begingroup$ Why is B=1/2? x-Ax-B=0, I understand why A=1 but why B=1/2? $\endgroup$– Olp112Jan 24 at 20:46
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$\begingroup$ 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. $\endgroup$ Jan 24 at 20:49
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$((x+1/2)^2-1/4)^{1/2}-Ax-B;$
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 $B=1/2.$
answered Jan 24 at 21:21