Maclauren series question...

Find the first three non-zero terms in the Maclauren series of:

f(x) = e^[tan^(-1)x] - 1

Hence, or otherwise, show that

(x->0) lim {e^[tan^(-1)x] - 1 - x - (x³/2)} / x³ = -1/6

NB: im not sure if its to the power of 3 in (x³/2) the writing is blurred. But i think thats it.

Note: Given the series for f(x), the power should be 2, not 3 in your limit...
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To find the series for f(x):

Start with the geometric series 1/(1 - x) = 1 + x + x^2 + ...

Let x = -t^2:
1/(1 + t^2) = 1 - t^2 + t^4 - ...

Integrate both sides from 0 to x:
arctan x = x - x^3/3 + x^5/5 - ...
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Therefore,
f(x) = e^(arctan x) - 1
.....= e^(x - x^3/3 + ...) - 1
.....= [1 + (x - x^3/3 + ...) + (x - x^3/3 + ...)^2/2! + (x - x^3/3 + ...)^3/3! + ...] - 1
.....= (x - x^3/3 + ...) + (1/2)(x - x^3/3 + ...)^2 + (1/6)(x - x^3/3 + ...)^3 + ...

Collecting the smallest degree terms yields
f(x) = x + (1/2) x^2 - (1/6)x^3 + ...
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Therefore,
lim(x→0) [e^(arctan x) - 1 - x - x^2/2] / x^3
= lim(x→0) [(x + (1/2) x^2 - (1/6)x^3 + ...) - x - x^2/2] / x^3
= lim(x→0) [(-1/6) x^3 + ...] / x^3
= lim(x→0) [-1/6 + (powers of x)]
= -1/6.

I hope this helps!