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(-tan(x)+sin(x))/x^3
Limit of the function/ (-tan(x)+sin(x))/x^3

Limit of the function (-tan(x)+sin(x))/x^3

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     /-tan(x) + sin(x)\
 lim |----------------|
x->0+|        3       |
     \       x        /
lim⁡x→0+(sin⁡(x)−tan⁡(x)x3)\lim_{x \to 0^+}\left(\frac{\sin{\left(x \right)} - \tan{\left(x \right)}}{x^{3}}\right) 
Limit((-tan(x) + sin(x))/x^3, x, 0)
Lopital's rule
We have indeterminateness of type
0/0,

i.e. limit for the numerator is
lim⁡x→0+(sin⁡(x)−tan⁡(x))=0\lim_{x \to 0^+}\left(\sin{\left(x \right)} - \tan{\left(x \right)}\right) = 0
and limit for the denominator is
lim⁡x→0+x3=0\lim_{x \to 0^+} x^{3} = 0
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
lim⁡x→0+(sin⁡(x)−tan⁡(x)x3)\lim_{x \to 0^+}\left(\frac{\sin{\left(x \right)} - \tan{\left(x \right)}}{x^{3}}\right)
=
lim⁡x→0+(ddx(sin⁡(x)−tan⁡(x))ddxx3)\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(\sin{\left(x \right)} - \tan{\left(x \right)}\right)}{\frac{d}{d x} x^{3}}\right)
=
lim⁡x→0+(cos⁡(x)−tan⁡2(x)−13x2)\lim_{x \to 0^+}\left(\frac{\cos{\left(x \right)} - \tan^{2}{\left(x \right)} - 1}{3 x^{2}}\right)
=
lim⁡x→0+(ddx(cos⁡(x)−tan⁡2(x)−1)ddx3x2)\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(\cos{\left(x \right)} - \tan^{2}{\left(x \right)} - 1\right)}{\frac{d}{d x} 3 x^{2}}\right)
=
lim⁡x→0+(−sin⁡(x)−2tan⁡3(x)−2tan⁡(x)6x)\lim_{x \to 0^+}\left(\frac{- \sin{\left(x \right)} - 2 \tan^{3}{\left(x \right)} - 2 \tan{\left(x \right)}}{6 x}\right)
=
lim⁡x→0+(ddx(−sin⁡(x)−2tan⁡3(x)−2tan⁡(x))ddx6x)\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(- \sin{\left(x \right)} - 2 \tan^{3}{\left(x \right)} - 2 \tan{\left(x \right)}\right)}{\frac{d}{d x} 6 x}\right)
=
lim⁡x→0+(−cos⁡(x)6−tan⁡4(x)−4tan⁡2(x)3−13)\lim_{x \to 0^+}\left(- \frac{\cos{\left(x \right)}}{6} - \tan^{4}{\left(x \right)} - \frac{4 \tan^{2}{\left(x \right)}}{3} - \frac{1}{3}\right)
=
lim⁡x→0+(−cos⁡(x)6−tan⁡4(x)−4tan⁡2(x)3−13)\lim_{x \to 0^+}\left(- \frac{\cos{\left(x \right)}}{6} - \tan^{4}{\left(x \right)} - \frac{4 \tan^{2}{\left(x \right)}}{3} - \frac{1}{3}\right)
=
−12- \frac{1}{2}
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 3 time(s)
The graph
02468-8-6-4-2-10105-10
Plot the graph
Rapid solution [src] 
-1/2
−12- \frac{1}{2}
Expand and simplify
Other limits x→0, -oo, +oo, 1
lim⁡x→0−(sin⁡(x)−tan⁡(x)x3)=−12\lim_{x \to 0^-}\left(\frac{\sin{\left(x \right)} - \tan{\left(x \right)}}{x^{3}}\right) = - \frac{1}{2}
More at x→0 from the left
lim⁡x→0+(sin⁡(x)−tan⁡(x)x3)=−12\lim_{x \to 0^+}\left(\frac{\sin{\left(x \right)} - \tan{\left(x \right)}}{x^{3}}\right) = - \frac{1}{2}
lim⁡x→∞(sin⁡(x)−tan⁡(x)x3)\lim_{x \to \infty}\left(\frac{\sin{\left(x \right)} - \tan{\left(x \right)}}{x^{3}}\right)
More at x→oo
lim⁡x→1−(sin⁡(x)−tan⁡(x)x3)=−tan⁡(1)+sin⁡(1)\lim_{x \to 1^-}\left(\frac{\sin{\left(x \right)} - \tan{\left(x \right)}}{x^{3}}\right) = - \tan{\left(1 \right)} + \sin{\left(1 \right)}
More at x→1 from the left
lim⁡x→1+(sin⁡(x)−tan⁡(x)x3)=−tan⁡(1)+sin⁡(1)\lim_{x \to 1^+}\left(\frac{\sin{\left(x \right)} - \tan{\left(x \right)}}{x^{3}}\right) = - \tan{\left(1 \right)} + \sin{\left(1 \right)}
More at x→1 from the right
lim⁡x→−∞(sin⁡(x)−tan⁡(x)x3)\lim_{x \to -\infty}\left(\frac{\sin{\left(x \right)} - \tan{\left(x \right)}}{x^{3}}\right)
More at x→-oo
One‐sided limits [src] 
     /-tan(x) + sin(x)\
 lim |----------------|
x->0+|        3       |
     \       x        /
lim⁡x→0+(sin⁡(x)−tan⁡(x)x3)\lim_{x \to 0^+}\left(\frac{\sin{\left(x \right)} - \tan{\left(x \right)}}{x^{3}}\right) 
-1/2
−12- \frac{1}{2} 
= -0.5
     /-tan(x) + sin(x)\
 lim |----------------|
x->0-|        3       |
     \       x        /
lim⁡x→0−(sin⁡(x)−tan⁡(x)x3)\lim_{x \to 0^-}\left(\frac{\sin{\left(x \right)} - \tan{\left(x \right)}}{x^{3}}\right) 
-1/2
−12- \frac{1}{2} 
= -0.5
= -0.5
Numerical answer [src] 
-0.5
-0.5
The graph
Limit of the function (-tan(x)+sin(x))/x^3
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