Finding Limits: ln and L'hopital's Rule Explained

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SUMMARY

The limit of the expression lim[p→0] ln(2x^p + 3y^p)/p^2 does not require L'Hôpital's Rule, as the numerator approaches a constant value rather than an indeterminate form. The discussion emphasizes the importance of analyzing the behavior of the logarithmic function as p approaches 0, particularly noting that the limit can vary based on the values of x and y, which must be non-negative for the limit to exist. A rigorous approach involves bounding the logarithmic expression for small p to clarify the limit's behavior.

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  • Understanding of limits in calculus
  • Familiarity with L'Hôpital's Rule
  • Knowledge of logarithmic functions and their properties
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Students studying calculus, particularly those focusing on limits and logarithmic functions, as well as educators seeking to clarify concepts related to L'Hôpital's Rule and limit evaluation techniques.

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Homework Statement



lim[p\rightarrow0] \frac{ ln(2x^{p}+3y^{p})}{p^2}

Homework Statement




I first plugged in zero to see if i can you L'hopital's rule
I got a (constant)/0 which is not an indeterminant form.

If I can't use the rule, what should be my next step to find the limit?
( I kinda guessed that the the answer's infinity but I'm not sure)
 
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You don't need L'hopital's rule here since the numerator is always constant... The answer is more straight forward than you think...

What's lim [x->infinity] 1/(x^2) ?

edit: didn't notice that x was to the power of p...see office shredder's comment
 
Last edited:
The numerator isn't constant, but it approaches one...
also, the limit is as p goes to 0, not infinity.

I suspect the easiest way to be rigorous is to bound ln(...) from above and below for small enough p, which gives you that constant/0 term. It's not always infinity though... you haev to be careful of the existence and sign of ln(...) depending on what values x and y take (note they have to be non-negative, or the limit definitely doesn't exist, for example)
 

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