Limit&nbsp;limx→0{tan⁡{x}−sin⁡{x}x3}So what I tried is:limx→0sin⁡{x}cos⁡{x}−sin⁡{x}x3=limx→0{sin⁡{x}}(1cos⁡x−1)x⋅x2From here, using the rule&nbsp;limx→0sin⁡{x}x=1&nbsp;it remains to evaluatelimx→01cos⁡{x}−1x2&nbsp;

From your last line,1cos⁡{x}−1x2=1−cos2(x)cos⁡(x)(1+cos⁡(x))x2=1cos⁡(x)(1+cos⁡(x))⋅(sin⁡(x)x)2

tan⁡x−sin⁡xx3=tan⁡xx1−cos⁡xx2=tan⁡xx2sin2x2x2=12tan⁡xx(sin⁡x2x2)2→12

From your last line,1cos⁡{x}−1x2=1−cos2(x)cos⁡(x)(1+cos⁡(x))x2=1cos⁡(x)(1+cos⁡(x))⋅(sin⁡(x)x)2

tan⁡x−sin⁡xx3=tan⁡xx1−cos⁡xx2=tan⁡xx2sin2x2x2=12tan⁡xx(sin⁡x2x2)2→12

"Limitlimx→0{tan⁡{x}−sin⁡{x}x3}So what I tried is:limx→0sin⁡{x}cos⁡{x}−sin⁡{x}x3=limx→0{sin⁡{x}}(1cos⁡x−1)x⋅x2From here, using the..." was answered by students like you

dotzis16xd

Answered question

2022-03-21

Limit $lim_{x \to 0} {\frac{\tan {x} - \sin {x}}{x^{3}}}$
So what I tried is:
$lim_{x \to 0} \frac{\frac{\sin {x}}{\cos {x}} - \sin {x}}{x^{3}} = lim_{x \to 0} \frac{{\sin {x}} (\frac{1}{\cos x} - 1)}{x \cdot x^{2}}$
From here, using the rule $lim_{x \to 0} \frac{\sin {x}}{x} = 1$ it remains to evaluate
$lim_{x \to 0} \frac{\frac{1}{\cos {x}} - 1}{x^{2}}$

Answer & Explanation

Ronald Martinez

Beginner2022-03-22Added 7 answers

From your last line,

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

Jonathon Hanson

Beginner2022-03-23Added 9 answers

\frac{\tan x - \sin x}{x^{3}} = \frac{\tan x}{x} \frac{1 - \cos x}{x^{2}} = \frac{\tan x}{x} \frac{2 \sin^{2} \frac{x}{2}}{x^{2}} = \frac{1}{2} \frac{\tan x}{x} {(\frac{\sin \frac{x}{2}}{\frac{x}{2}})}^{2} \to \frac{1}{2}

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