Verified answer

Given: lim(x-->infinity) [(x²+1) arcsin(2x/(x²+1)) - 2x]

Let's rewrite it as follows:

lim(x-->infinity) [arcsin(2x/(x²+1)) - 2x/(x²+1)] / (x²+1)^(-1).

Since this is of the form "0/0", apply L'Hopital's Rule:

lim(x-->infinity) {1/(sqrt[1 - (2x/(x²+1))²]) - 1} {(2 - 2x²)/(x²+1)²} /

[-2x /(x²+1)²].

= lim(x-->infinity) {[1 - (2x/(x²+1))²]^(-1/2) - 1} / {x / (x² - 1)}.

Again, this is of the form "0/0"; so apply L'Hopital's Rule again.

Carefully taking derivatives, we get

lim(x-->infinity) (-1/2) * [1 - (2x/(x² + 1))²]^(-3/2) * 2 * (2x/(x²+1))^1 *

{(2 - 2x²)/(x² + 1)²} / {-(x² + 1) / (x² - 1)²}

= lim(x-->infinity) -[1 - (2x/(x² + 1))²]^(-3/2) * {4x (x² - 1)^3 / (x² + 1)^4}

= - (1 - 0)^(-3/2) * 0

= 0.

I hope this helps!

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Notes:

(1) I verified this answer three ways: Wolfram Alpha, my graphing calculator, and power series...

(2) The algebra should be pretty much intact above, although it is easy to make a typo with the complicated expressions above. Good luck with your class!

I like it better with series. I think it is useful to see this solution here.

The first thought was to use known fact that arcsint ~ t, t ->0 because 2x/(x² + 1) tends to 0 as x-> ∞.

But in this case everything simplifies to 0. That means it is not enough to substitute arcsint with t, we need more terms from arcsine's power series.

arcsint = t + (1/6) t³ + (3/40) t^5 + (5/112) t^7 + ..., |t| <1.

Let's take two of them:

arcsint ~ t + t³/6 as t->0 and substitute t = 2x/(x²+1)

lim (x²+1)·[2x/(x²+1) + {2x/(x²+1)}³/6] - 2x = lim (x²+1)·[2x/(x²+1) + 8x³/{6(x²+1)³}] - 2x =

= lim 2x+ (4/3)·x³/(x²+1)² - 2x = lim (4/3)·x³/(x²+1)² = lim(4/3)·x³/(x^4 + 2x²+ 1) =

= lim(4/3)·x³/[(x^4)(1 + 2/x²+ 1/(x^4))] = lim(4/3)/[x(1 + 2/x²+ 1/(x^4))] = 0 as x-> ∞

assuming parentheses are correct, lim is infinity