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Claim: e^{t}= 1 + t +o(t)

Proof:

e^{t}= 1 + t + t^{2}/2! + t^{3}/3! +...

Let g(t)=t^{2}/2! + t^{3}/3! +...

g(t) is o(t), thus e^{t}= 1 + t +o(t).

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I don't understand why g(t) is o(t).

Why is it true that

lim g(t)/t = 0 ?

t->0

I know that for example lim[f(x)+g(x)]=lim f(x) + lim g(x), so if lim f(x)=0 and lim g(x)=0, then

lim[f(x)+g(x)] = 0+0 = 0. I believe this property is true only when computing the limit of a sum of a FINITE number of functions.

But g(t)/t above is an infinite sum; it has an infinite number of terms. How can we compute and prove that the limit of g(t)/t is 0? I know each term goes to 0, but we are summing an INFINITE number of terms, so how can you be sure that the limit is 0?

Any help/explanations would be much apprecaited!

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# Little o function

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