State the value of x for which the
Series is convergent
X/1 - x²/2+ x/³/3 - x/4⁴
Answers
Answer:
Mat104 Solutions to Taylor and Power Series Problems from Old Exams
(1) (a). This is a 0/0 form. We can use Taylor series to understand the limit.
e
x = 1 + x +
x
2
2! +
x
3
3! +
x
4
4! + · · · +
x
n
n!
+ . . .
e
−x = 1 − x +
x
2
2! −
x
3
3! +
x
4
4! · · · +
(−1)nx
n
n!
+ . . .
Thus e
x − e
−x = 2x +
2x
3
3! +
2x
5
5! + . . .
From this we find that
e
x − e
−x − 2x =
2x
3
3! + higher degree terms
As x approaches 0, the lowest power of x will dominate because the higher degree terms
vanish much more rapidly. We can say that
e
x + e
−x − 2x ∼
2x
3
3! as x → 0.
Next we consider the denominator.
x ln(1 + x) = x(x − x
2
/2 + x
3
/3 − x
4
/4 + . . .) = x
2 −
x
3
2
+
x
4
3
−
x
5
4
+ . . . .
Thus the denominator x
2 − x ln(1 + x) will be dominated by its lowest degree term x
3
2
as
we let x → 0 and so
e
x + e
−x − 2x
x
2 − x ln(1 + x)
∼
2x
3/3!
x
3/2
=
4
6
=
2
3
as x → 0.
(1b) Again we have a 0/0 form. In a similar manner we manipulate Taylor series to
determine what power of x the numerator and denominator resemble as x approaches 0.
First recall that
cos x = 1 − x
2
/2! + x
4
/4! − x
6
/6! + . . . and sin x = x − x
3
/3! + x
5
/5! − x
7
/7! + . . .
Then we can easily compute that
cos x
2 − 1 + x
4/2 = x
8/4! plus higher degree terms
x
2
(x − sin x)
2 = x
8/(3!3!) plus higher degree terms
Thus
cos x
2 − 1 + x
4/2
x
2
(x − sin x)
2
∼
x
8/4!
x
8/(3!3!) =
3!3!
4! =
3
2
as x → 0
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