## Taylor Expansion of Natural Logarithm

That's why it is undetermined and not infinity.
x=50 is the smallest x for which the limit will always be 0.
anything lower than 50, and the nominator could be larger than the denominator => the limit will diverge.
 Mentor "It could be larger" => you cannot use the general version of the lagrangian error to show convergence It converges, but you cannot use this way to show it. With c arbitrary small, it would diverge for other x as well, by the way.
 yes, but since c is between x and 100, if x is bigger than 50 or smaller than 200 it won't diverge for any x and for any c. when i pick x=1, for example, for 1
 Mentor Let me rewrite your limit a bit: $$\lim_{n \to \infty} \frac{1}{c(n+1)} \frac{(x-100)^n}{c^n}$$ Consider c=100-epsilon with some (small) constant epsilon: The first part goes to 0, the second part, neglecting its sign, can be written as $$\frac{(100-x)^n}{(100-\epsilon)^n}$$ This is smaller than 1 for epsilon
 It's very confusing. If the Lagrange remainder doesn't approach 0 it means you cannot estimate the function correctly, but in this case the calculation of the series does give accurate estimations. It's very interesting. Does anyone know the answer to this?
 Recognitions: Homework Help ^The Lagrange remainder does approach 0, you have chosen a terrible way to estimate it. Either use a more subtle estimate of the Lagrange remainder or use the Cauchy Remainder or Schlömilch Remainder. Remember your estimate for c is quite crude.

 Quote by lurflurf ^The Lagrange remainder does approach 0, you have chosen a terrible way to estimate it. Either use a more subtle estimate of the Lagrange remainder or use the Cauchy Remainder or Schlömilch Remainder. Remember your estimate for c is quite crude.
The remainder must approach 0 for every c between x and 100.
If i can find even 1 such number for which the remainder does not approach 0, it means that the series is not a correct estimation for the function.

Mentor
 Quote by golanor If i can find even 1 such number for which the remainder does not approach 0, it means that the series is not a correct estimation for the function.
No: If you cannot find even 1 sequence cn, ...
But I found one.

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