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Limit with Taylor Series

  • #1

Homework Statement


[tex] \lim_{x \to 0}[\frac{\sin(\tan(x))-\tan(\sin(x))}{x^7}][/tex]


Homework Equations


[tex]\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!} + ...[/tex]
[tex]\tan(x)=x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{215}+ ...[/tex]


The Attempt at a Solution


I have an idea of how to do this by replacing sin(tan(x)) with tan(x) - tan(x)^3/3! + tan(x)^5/5!, etc., and then replacing the tans and sines with their respective taylor series. But I'm supposed to be able to do this without a calculator and presumably without expanding polynomials to the seventh power, which seems a bit ridiculous.

Can someone point me in the right direction?
 

Answers and Replies

  • #2
vela
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You essentially want to do what you said: expand both functions as Taylor series, multiply everything out, and then simplify, but, in practice, you don't actually multiply everything out. You just need to keep track of what's going to contribute to each term.

For example, consider the ##x^3## term in the expansion of ##\sin(\tan x)##. Contributions to it come from the combination of the linear term in the expansion of tan x and the x3 of sin x, or from the combination of the x3 term of tan x and the linear term of sin x. So the x^3 term will be
$$\frac{x^3}{3} - \frac{x^3}{3!} = \frac{x^3}{6}$$
 

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