Could somebody help me with this problem?
Find the Taylor polynomials the 2nd Taylor polynomial around 0 for the following function
f(x) = (1)/(1+x)
and this nasty question...
Using l'Hoptal's rule...
lim (x->infin) [((cos) (a/x))^x^2] where a>0
lim (x->0) [((log(1+ax))/x) where a>0
Hi Fant
Just realised the first part was simple apparently so I got the following with few other examples:
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Using l'hopital rule to compute limits
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lim x->0 (1 - cos x)/ x^2
lim x->0 (sin x)/ 2x
lim x->0 (sin x)/ x = 1
lim x->0 (cos x)/ 2 = 1/2
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lim x->0 (a^x - 1)/ k*x
lim x->0 [xa^(x-1) - 1]/ k = 1
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And the ending of
lim x->infin (cos (a/x))^(x^2)
Is the limit 1/2?
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Now the Taylor Polynomial
f(x) = (1)/(1+x)
f'(x) = (-1)(1+x)^-2
f''(x) = (-2)(1+x)^-\3
f(0) = 1
f'(0) = -1
f''(0) = -2
Which will gave me (1/0!)-(1/1!)x-(2/2!)x^2 = 1-x-x^2
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How did I do?