Compute the limit by.....?

Question

im [log(1-x) + x*e^(x/2)] / x^3

x->0

and

lim log(1+x^2) / 2x

x-->0

using the Taylor Polynomials

I can do them with L'Hopitals but when you use the Taylor series, how do you know how many terms to keep?

Answer 1

1/(1+w) = Σ (-1)^n w^n

log(1+w) is the integral .. . .

thus

log(1 + w) = Σ (-1)^n w^(n+1) / (n+1)

meanwhile

e^w = Σ w^n / n!

thus

log(1-x) = -Σ x^(n+1) / (n+1)

xe^(x/2) = x* Σ (x/2)^n / n!

log(1-x) + x*e^(x/2) = Σ x^(n+1) * [-1/(n+1) + 1/(2^n n!)]

if you expand this... you will observe that it starts actually at the cubic.

the coefficient is -5/24 (this is the value of the limit)

the second question

log(1 + x^2) = Σ (-1)^n (x^2)^(n+1) / (n+1)

log(1 + x^2)/2x = Σ (-1)^n x^(2n+1) / 2(n+1)

thus log(1 + x^2)/2x = (x / 2)  - (x^3 / 4) + (x^5 / 6) + ...

and as x goes to 0,

the whole thing is 0.

Answer 2

Have you ever listened to Elgar's Variations? or work by Litolff...he has a scherzo when played on a piano is very uplifting.{looked at some of your answers}...also f(x) = ( x² + x-2) /( x-1) does not have a vertical asymptote at x = 1 while g(x) = f(x) /(x-1) does..do you know why?...as to this problem the terms left need to be O(x^4) so that after the division by x^3 all the remaining terms have an x in them and thus tend to zero, likewise for the second O(x²)...remember that O(h) means all terms have an ' h ' in them. Keep up the good questions