Indeterminate Products Giving Me Two Different Limits

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SUMMARY

The limit of the function x ln(x) as x approaches 0 is evaluated using L'Hôpital's Rule, yielding a result of 0. The user initially misapplies the rule, leading to confusion about the limit's behavior. The correct application involves differentiating both the numerator and denominator appropriately, confirming that the limit converges to 0. This discussion highlights common pitfalls in calculus, particularly in limit evaluation.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with L'Hôpital's Rule
  • Basic knowledge of logarithmic functions
  • Ability to differentiate functions
NEXT STEPS
  • Study the application of L'Hôpital's Rule in various limit scenarios
  • Explore the properties of logarithmic functions and their limits
  • Practice differentiating complex functions to reinforce calculus skills
  • Review common mistakes in limit evaluations to improve accuracy
USEFUL FOR

Students studying calculus, educators teaching limit concepts, and anyone looking to strengthen their understanding of L'Hôpital's Rule and logarithmic limits.

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Homework Statement



lim x-->0 of [tex]x lnx[/tex]

Homework Equations


The Attempt at a Solution



1) [tex]\frac{lnx}{1/x}[/tex] = [tex]\frac{1/x}{-1/x^{2}}[/tex] = (-x) = 0

2) [tex]\frac{x}{1/lnx}[/tex] = [tex]\frac{1}{1/1/x}[/tex] = [tex]\frac{1}{x}[/tex] = [tex]\infty[/tex]
 
Last edited:
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hmmm...are you sure [tex]\frac{d}{dx} \frac{1}{\ln (x)}=\frac{1}{\frac{1}{x}}[/tex]?:wink:
 
yea I realized it after a minute of looking at it again. this is exactly the sort of stupid mistake that lowers my marks in tests.

and oops. I didn't see you answer or I wouldn't have edited the entry back.

edit edit: there, I put it back up... but for some reason I think I made a double of the thread... agh. it's 4 AM. I'm tired :smile:

edit edit edit: and, of course: thanks.

(I go sleep now)
 
Last edited:

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