Limit of Function with Log: What is the Limit of f(x) as x Approaches Infinity?

MaxManus
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Homework Statement



Find the limit of
f(x) = x^2 ln(1+1/x) - x

x-> ∞




The Attempt at a Solution



Not sure where to start, but I take the derivative

f'(x) = 2x*ln(1+1/x) +x^2 \frac{ -x^{-2}}{1+1/x} -1
2x*ln(1+1/x) - \frac{1}{1+1/x} -1

the second term goes to -1 as x->∞ ant the last term is always -1. Can I say somethong about the last term and will it help me?
 
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Got it.

f(x) = x^2 ln(1+1/x) - x

= \frac{ln(1+1/x) - (1/x)}{1/x^2}

u = 1/x

f(u) = (ln(1+u) -u)/u^2

which limit is -1/2 when u->0

The limit is found by using l hopital twice
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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