We have an answer to "Limit limlimits{x→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..."

Eileen Garza

Answered question

2022-04-28

Limit

lim_{lim} i t s {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}}

haillarip0c9

Beginner2022-04-29Added 23 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}

Ezakhenitne

Beginner2022-04-30Added 10 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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