What is taylor 1/(x-1),6 ?

The answer to taylor 1/(x-1),6 is 1/5-1/25 (x-6)+1/125 (x-6)^2-1/625 (x-6)^3+1/3125 (x-6)^4+\ldots

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taylor 1/(x-1),6

taylor

\frac{1}{x - 1}, 6

\frac{1}{5} - \frac{1}{25} (x - 6) + \frac{1}{125} (x - 6)^{2} - \frac{1}{625} (x - 6)^{3} + \frac{1}{3125} (x - 6)^{4} + \dots

Solution steps

Apply the Taylor Formula

= \frac{1}{5} + \frac{\frac{d}{d x} ( \frac{1}{x - 1} ) ( 6 )}{1 !} (x - 6) + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{1}{x - 1} ) ( 6 )}{2 !} (x - 6)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{1}{x - 1} ) ( 6 )}{3 !} (x - 6)^{3} + \dots

Evaluate Derivatives

= \frac{1}{5} + \frac{- \frac{1}{25}}{1 !} (x - 6) + \frac{\frac{2}{125}}{2 !} (x - 6)^{2} + \frac{- \frac{6}{625}}{3 !} (x - 6)^{3} + \frac{\frac{24}{3125}}{4 !} (x - 6)^{4} + \dots

Refine

= \frac{1}{5} - \frac{1}{25} (x - 6) + \frac{1}{125} (x - 6)^{2} - \frac{1}{625} (x - 6)^{3} + \frac{1}{3125} (x - 6)^{4} + \dots

x \to 0 lim (\frac{e ^{x} - 1}{18 x})

\frac{x}{2 x + x - 3}

\int (in x)^{2} d x

x \to \infty lim (1 + x)

y = 2 (x + 1), y = 3 (x + 1), x = 5

What is taylor 1/(x-1),6 ?
The answer to taylor 1/(x-1),6 is 1/5-1/25 (x-6)+1/125 (x-6)^2-1/625 (x-6)^3+1/3125 (x-6)^4+\ldots