Use Taylor Series To Evaluate

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SUMMARY

The discussion focuses on using Taylor series to evaluate the limit of \(\frac{\ln(x)}{(x-1)}\) as \(x\) approaches 0. Participants suggest that the limit may be more relevant as \(x\) approaches 1, prompting the use of the Taylor series expansion for \(\ln(1+(x-1))\). The correct approach involves dividing the Taylor series by \(x-1\) to find the limit, leading to a conclusion that the limit evaluates to 1.

PREREQUISITES
  • Understanding of Taylor series expansions, specifically for \(\ln(1+x)\) and \(\ln(1-x)\)
  • Knowledge of limits in calculus, particularly evaluating limits as \(x\) approaches specific values
  • Familiarity with logarithmic functions and their properties
  • Basic algebraic manipulation skills for simplifying expressions
NEXT STEPS
  • Study the Taylor series expansion for \(\ln(1+x)\) and \(\ln(1-x)\)
  • Learn techniques for evaluating limits using L'Hôpital's Rule
  • Explore the concept of indeterminate forms in calculus
  • Practice problems involving limits and Taylor series in calculus textbooks
USEFUL FOR

Students studying calculus, particularly those focusing on limits and Taylor series, as well as educators seeking to clarify these concepts for their students.

xtrubambinoxpr
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1. Homework Statement [/b]
use taylor series to evaluate lim x -> 0 of \frac{ln(x)}{(x-1)}


Homework Equations



I know that -ln (1-x) taylor polynomial
and that of ln (1+x)

The Attempt at a Solution



Using the basics that I know I would assume I would just make ln (1+x) = ln (x) by making x = x-1

so ln (1+(x-1)) = ln x

But i don't know if that is correct
 
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xtrubambinoxpr said:
1. Homework Statement [/b]
use taylor series to evaluate lim x -> 0 of \frac{ln(x)}{(x-1)}


Homework Equations



I know that -ln (1-x) taylor polynomial
and that of ln (1+x)

The Attempt at a Solution



Using the basics that I know I would assume I would just make ln (1+x) = ln (x) by making x = x-1

so ln (1+(x-1)) = ln x

But i don't know if that is correct


Why would you need Taylor series for ##\lim_{x\to 0}##? I guess maybe you really mean ##x\to 1##, which is more interesting? If so, do what you suggest by writing the series for ##\ln(1+(x-1)## and divide by ##x-1##.
 
LCKurtz said:
Why would you need Taylor series for ##\lim_{x\to 0}##? I guess maybe you really mean ##x\to 1##, which is more interesting? If so, do what you suggest by writing the series for ##\ln(1+(x-1)## and divide by ##x-1##.

correct me if I am wrong but I got 1
 

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