Use Taylor Series To Evaluate

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
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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