Get the solution to "limx→0(11−cos⁡x−2x2)"

Jaylin Clements

Answered question

2022-03-31

lim_{x \to 0} (\frac{1}{1 - \cos x} - \frac{2}{x^{2}})

armejantm925

Beginner2022-04-01Added 20 answers

Let x=2y

lim_{x \to 0} (\frac{1}{1 - \cos x} - \frac{2}{x^{2}})

= \frac{1}{2} lim_{y \to 0} (\frac{1}{\sin^{2} y} - \frac{1}{y^{2}})

= \frac{1}{2} lim_{y \to 0} (1 + \frac{\sin y}{y}) lim_{y \to 0} (\frac{y - \sin y}{y^{3}}) {(lim_{y \to 0} \frac{y}{\sin y})}^{2}

Now use Are all limits solvable without L'Hôpital Rule or Series Expansion for the middle limit

kattylouxlvc

Beginner2022-04-02Added 11 answers

\frac{1}{1 - \cos x} - \frac{2}{x^{2}} = \frac{1}{2 \sin^{2} (\frac{x}{2})} - \frac{2}{x^{2}}

= \frac{\frac{x^{2}}{4}}{\sin^{2} (\frac{x}{2})} \times \frac{2}{x^{2}} - \frac{2}{x^{2}}

= ({(\frac{\frac{x}{2}}{\sin (\frac{x}{2})})}^{2} - 1) \frac{2}{x^{2}}

But

\frac{x}{\sin x} = 1 + \frac{x^{2}}{6} + o (x^{3})

and therefore

{(\frac{\frac{x}{2}}{\sin (\frac{x}{2})})}^{2} = 1 + \frac{x^{2}}{12} + o (x^{3})

Therefore, your limit is

\frac{2}{12} = \frac{1}{6}

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