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How would I calculate $$\lim_{x \rightarrow 0}{\frac{\int_{0}^{x} \frac{\sin(t)}{t}-\tan(x)}{2x(1-\cos(x))}}$$ using a Maclaurin polynomial? For the integral in the numerator, the polynomial doesn't seem to exist beyond degree 1.

Thanks!

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2 Answers

up vote 1 down vote accepted

There is no trouble with the integral. Just write down the expansion of $\sin t$, divide by $t$, and integrate term by term.

In detail, $\sin t$ has expansion $t-t^3/6+\cdots$. Divide by $t$. We get $1-t^2/6+\cdots$. Integrate from $0$ to $x$. We get $x-x^3/18+\cdots$. If you are worried about what happens at $0$, don't. The limit is $1$. Anyway, the integral is not sensitive to the value of the function at just one point.

You will need a bit of the expansion of $\tan x$. Possibly you could just compute the terms (you will need up to the $x^3$ term) from the definition. You will need the first three derivatives of $\tan x$. There are other ways, like working with $\sin x$ and $\cos x$.

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I just realized why I was confused. Computed a certain part wrong. Thanks! –  ro44 Feb 24 '12 at 1:25
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We have, for $x$ near zero, $$ \sin t = t-t^3/6+o(t^5), \ \ \tan x =x+x^3/3+o(x^5),\ \ \cos x =1-x^2/2+o(x^4). $$ Then $$ \frac{\int_0^x\frac{\sin t}t dt-\tan x}{2x(1-\cos x)} =\frac{\int_0^x(1-t^2/6+o(t^4))dt-(x+x^3/3+o(x^5))}{2x(x^2/2+o(x^4))} =\frac{x-x^3/18+o(x^5)-(x+x^3/3+o(x^5))}{2x(x^2/2+o(x^4))} =\frac{-7x^3/18+o(x^5)}{x^3+o(x^5))} =\frac{-7/18+o(x^2)}{1+o(x^2))} \to -\frac7{18}. $$

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