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

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SUMMARY

The limit of the function f(x) = x^2 ln(1+1/x) - x as x approaches infinity is determined to be -1/2. The solution involves applying L'Hôpital's Rule twice after transforming the function into f(u) = (ln(1+u) - u)/u^2, where u = 1/x. The derivative calculations confirm that the second term approaches -1, leading to the conclusion that the overall limit is indeed -1/2.

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  • Knowledge of logarithmic functions
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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) = [itex]2x*ln(1+1/x) +x^2 \frac{ -x^{-2}}{1+1/x} -1[/itex]
[itex]2x*ln(1+1/x) - \frac{1}{1+1/x} -1[/itex]

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

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

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
 
Last edited:

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