So my question is how to find $$\lim_{x\to\infty}\left(x\left(\pi-2\arcsin\frac{x}{\sqrt{x^2+1}}\right) \right)$$
the answer should be 2
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good! What have you done? Where are you stuck? :) – Ant Mar 18 '15 at 18:00
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I really can not understand what to do the next step. – Mar 18 '15 at 18:01
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1Write $1/x$ as denominator. – Aaron Maroja Mar 18 '15 at 18:01
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2Hint: You have the indeterminate form $\infty\cdot 0$, you want either $\frac{\infty}{\infty}$ or $\frac{0}{0}$. @AaronMaroja gives a good complementary hint – jameselmore Mar 18 '15 at 18:02
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The differentiation is more pleasant if the arcsine is written as $\arctan x$. – André Nicolas Mar 18 '15 at 18:03
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So what to do after he brought in the denominator of x? – Mar 18 '15 at 18:05
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Use L'Hospital! – Aaron Maroja Mar 18 '15 at 18:06
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ok, now try.... – Mar 18 '15 at 18:07
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We are looking at $\frac{\pi-2\arctan x}{1/x}$. Differentiate top and bottom and simplify. We get $\frac{2x^2}{1+x^2}$. Now it's over. – André Nicolas Mar 18 '15 at 18:08
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@ André Nicolas wait.. and where the root of the missing? – Mar 18 '15 at 18:11
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I used the fact that $\arcsin(x/\sqrt{x^2+1})=\arctan x$. That got rid of the square root stuff. – André Nicolas Mar 18 '15 at 18:14
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I think about it :) – Mar 18 '15 at 18:17
2 Answers
1
As $\arcsin y+\arccos y=\dfrac\pi2$
We have $$F=2\lim_{x\to\infty^+}x\arccos\dfrac x{\sqrt{x^2+1}}$$
W/O L'Hôpital's rule,
Set $x=\cot y\implies x^2+1=\csc^2y$ $$F=2\lim_{y\to0^+}\dfrac y{\sin y}\cdot\cos y$$
Using L'Hôpital's rule, set $1/x=y$
$$F/2=\lim_{y\to0^+}\dfrac{\arccos\dfrac1{\sqrt{1+y^2}}}y$$
$$=\lim_{y\to0^+}\dfrac{-\dfrac1{\sqrt{1-1/(1+y^2)}}\cdot-\dfrac{2y}{2(1+y^2)^{3/2}}}1$$
$$=\lim_{y\to0^+}\dfrac{y}{y(1+y^2)^{3/2-1/2}}$$
as $\sqrt{y^2}=|y|$ which is $+y$ here as $y>0$
lab bhattacharjee
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1
$$F(x)=\left(x\left(\pi-2\arcsin\frac{x}{\sqrt{x^2+1}}\right) \right)$$ Putting $\frac{1}{x}$ to the denominator, we have $$F(x)=\frac{\left(\pi-2\arcsin\frac{x}{\sqrt{x^2+1}}\right)}{\frac{1}{x}}$$ The derivative of the numerator is $\frac{-2}{1+x^2}$ when $x\gt 0$ and the derivative of the denominator is $\frac{-1}{x^2}$ and therefore the limit is $2$
marwalix
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