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## Homework Statement

Evaluate the limit ##\lim_{x\to0} \frac{1}{x}(\frac{1}{tanx}-\frac{1}{x}) ## using Taylor's formula. (Hint: ##\frac{1}{1+c}=\frac{1-c^2+c^2}{1+c} ## may be useful)

## The Attempt at a Solution

I began by substituting ##tanx## with ##x+\frac{x^3}{3}+x^3ε(x)##, where ε tends to zero as x approaches 0.

##\frac{1}{x}(\frac{1}{tanx}-\frac{1}{x})=\frac{1}{x}(\frac{x-tanx}{xtanx})=\frac{1}{x}(\frac{x-x-\frac{x^3}{3}-x^3ε(x))}{x(x+\frac{x^3}{3}+x^3ε(x))})=\frac{1}{x}(\frac{-\frac{x^3}{3}-x^3ε(x))}{x(x+\frac{x^3}{3}+x^3ε(x))})=\frac{-\frac{x^3}{3}-x^3ε(x))}{x^2(x+\frac{x^3}{3}+x^3ε(x))}=\frac{-\frac{x^3}{3}-x^3ε(x))}{x^3+\frac{x^4}{3}+x^4ε(x))}=\frac{-\frac{1}{3}-ε(x))}{1+\frac{x}{3}+xε(x))}## →-1/3 as x→0

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