Calculating Limits: The Uncertainty of 0*Infinity

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SUMMARY

The discussion focuses on calculating the limit of the expression Lim (1+x)ln(e-1/x^2)-x as x approaches infinity. The user simplifies the expression to ln(e+1/x^2)+x(ln(e+1/x^2)-1), leading to a form of 1-(0*inf). The consensus is to apply L'Hôpital's Rule to resolve the indeterminate form, specifically transforming it into (ln(e+1/x^2)-1)/(1/x), which relates to the derivative of ln(e+u) at u=0. This approach clarifies the limit calculation without losing accuracy.

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Dell
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in a calculus problem i am doing, i need to find the Limit of

Lim (1+x)ln(e-1/x^2)-x
x->infinity

i have brought it to a point where i have ln(e+1/x^2)+x(ln(e+1/x^2)-1)
therefore my limit is 1-(0*inf)

can i leave this as 1-0=1 and that would be my answer or do i somehow need to change it around to get 0/0 of inf/inf??
 
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I am not sure how did you arrive on your second sentence, but from there, you need to use L'hopital Rule on (ln(e+1/x^2)-1)/(1/x), though if you change variables you get that this is definition of the derivative of the function ln(e+u) at the point u=0, have I ruined it for you, have I? :-)
 

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