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- Find each limit by writing the top and the bottom as series near 0 and keeping the lowest power that survives.\displaystyle\lim_{x \to 0} \dfrac{e^x - e^{-x} - 2x}{x^3}\displaystyle\lim_{x \to 0} \dfrac{\arctan x - \sin x}{x^3}\displaystyle\lim_{x \to 0} \dfrac{\cos x - e^{-x^2/2}}{x^4}\displaystyle\lim_{x \to 0} \dfrac{e^x - 1 - x}{x - \arctan x}
- The integral I = _0^0.74 (x³)\,dx has no antiderivative built from the familiar functions.
- Let $g
- Let $F
- Let $f
- Let $f
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Using series to evaluate limits and integrals: the lesson ›More Chapter 6 worksheets: Power series and Taylor series
- 6.1Taylor polynomials
- 6.2Power series and the interval of convergence
- 6.3Representing functions as power series
- 6.4The common Maclaurin series
- 6.5The Taylor remainder and the Lagrange error bound
- 6.6The binomial series
- 6.7Using series to evaluate limits and integrals