Evaluating the Limit of a Function at Infinity: [x+1-ln(x+1)]

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SUMMARY

The limit of the function as x approaches infinity, specifically lim x-> ∞ [x+1-ln(x+1)], can be evaluated using L'Hopital's rule after transforming the expression. By multiplying the numerator and denominator by the conjugate (x+1) + ln(x+1), the expression simplifies to ((x+1)^2 - (ln(x+1))^2) / ((x+1) + ln(x+1)), which results in the indeterminate form ∞/∞. This allows for the application of L'Hopital's rule to find the limit definitively.

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Homework Statement



[tex]\lim x-> \infty [x+1-ln(x+1)][/tex]

Homework Equations





The Attempt at a Solution


How does one evaluate this? I don't know how to use L'Hopital's rule on this and I have infinity- infinity, which is indeterminate. Thanks!
 
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There is a fairly standard technique for going from "a-b" that leads to "[itex]\infty[/itex]- [itex]\infty[/itex]" to a "0/0" or "[itex]\infty/\infty[/itex]", given in every Calculus text I know: multiply both numerator and denominator by the "conjuate" a+ b. In this case that gives
[tex]\frac{(x+1)^2- (ln(x+1))^2}{x+1+ ln(x+1)}[/tex]
That is now of the form "[itex]\infty/\infty[/itex]" and you can use L'Hopital's rule.
 
Ok thanks.
 

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