Calculus 2

Calculus Level 4

lim ⁡ x → 0 cos ⁡ x − cos ⁡ x 3 sin ⁡ 2 x \large \lim_{x\to0} \dfrac{ \sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2 x}

The limit above has a closed form. Find the value of this closed form.

Give your answer to 3 decimal places.


The answer is -0.0833.

Radhesh Sarma by Radhesh Sarma , 22, India

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1 solution

Rishabh Jain
Jun 15, 2016

Call the limit W \mathfrak W . Use sin ⁡ 2 x = 1 − cos ⁡ 2 x \color{#3D99F6}{\small{\sin^2 x=1-\cos^2x}} followed by substituting cos ⁡ x = t 6 \color{#3D99F6}{\small{\cos x=t^6}} .

W = lim ⁡ t → 1 t 3 − t 2 ⏞ t 2 ⏞ 1 ( t − 1 ) 1 − t 12 = − lim ⁡ t → 1 1 ( 1 − t 12 1 − t ) \large \mathfrak W=\displaystyle\lim_{t\to 1}\dfrac{\overbrace{t^3-t^2}^{\overbrace{t^2}^1(t-1)}}{1-t^{12}}=-\displaystyle\lim_{t\to 1}\dfrac{1}{\left(\color{#D61F06}{\frac{1-t^{12}}{1-t}}\right)}

Using sum of GP denominator (i.e 1 − t 12 1 − t \color{#D61F06}{\frac{1-t^{12}}{1-t}} ) is equal to 1 + t + t 2 + ⋯ + t 11 \color{#D61F06}{1+t+t^2+\cdots+t^{11}} . As t → 1 t\to 1 this tends to 12 12 ( Alternately you can use Lhopital Rule to get − 12 t 11 − t \dfrac{-12t^{11}}{-t} ) . Hence,

W = − 1 12 ≈ − 0.833 \large \mathfrak W=\frac{-1}{12}\approx \boxed{\color{#20A900}{-0.833}}

9 Helpful 0 Interesting 0 Brilliant 0 Confused

Without doing any substitution , you can directly use L hospital's Rule .

lim ⁡ x → 0 cos ⁡ x − cos ⁡ x 3 sin ⁡ 2 x lim ⁡ x → 0 1 2 ( cos ⁡ x ) 1 2 ⋅ − sin ⁡ x − 1 3 ( cos ⁡ x ) 2 3 ⋅ − sin ⁡ x 2 ⋅ sin ⁡ x ⋅ cos ⁡ x ⟹ 1 3 − 1 2 2 = − 1 12 \lim_{x \to 0}\dfrac{\sqrt{\cos{x}}-\sqrt[3]{\cos{x}} }{\sin^2{x}} \\\lim_{x \to 0}\dfrac{\dfrac{1}{2(\cos{x})^\frac12}\cdot - \cancel{\sin{x}}-\dfrac{1}{3(\cos{x})^{\frac23}}\cdot - \cancel{\sin{x}} }{2\cdot \cancel{\sin{x}}\cdot\cos{x}} \\ \implies \dfrac{\dfrac13-\dfrac12}{2} = \boxed{-\dfrac1{12}}

Sabhrant Sachan - 5 years ago

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Nice.... Obviously we can always use Lhopital rule directly on 0 / 0 0/0 forms but in this case using Lhopital directly will require pen and paper ( at least for me) ..... But I was so lazy to grab one so... :-)

Rishabh Jain - 5 years ago

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I wonder you did that substitution in your mind.

Akshay Yadav - 4 years, 12 months ago

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@Akshay Yadav – And that's why I bothered to write a 'non pen and paper' solution.. :-)

Rishabh Jain - 4 years, 12 months ago

There is one slight error, sin ⁡ 2 ( x ) = 1 − cos ⁡ 2 ( x ) \sin^2(x)=1-\cos^2(x) .

Akshay Yadav - 4 years, 12 months ago

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Thanks... Corrected

Rishabh Jain - 4 years, 12 months ago

I missed because of -tive sign.
L e t C o s 1 / 6 x = t . ⟹ L = lim ⁡ x → 0 cos ⁡ x − cos ⁡ x 3 sin ⁡ 2 x = lim ⁡ t → 0 t 3 − t 2 1 − t 12 = lim ⁡ t → 0 t 2 ( t − 1 ) 1 − t 12 B u t 1 − t 12 = ( 1 + t 6 ) ( 1 + t 3 ) ( 1 − t ) ( 1 + t + t 2 ) . ∴ L = 1 ∗ ( − 1 ) 2 ∗ 2 ∗ 1 ∗ 3 = − 1 12 = − 0.833333. Let~Cos^{1/6}x=t.~~ \\ \implies~L=\large \lim_{x\to0} \dfrac{ \sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2x}\\ =\large \lim_{t\to0} \dfrac{ t^3 - t^2}{1-t^{12}}= \lim_{t\to0} \dfrac{ t^2(t - 1)}{1-t^{12}} \\ But~1- t^{12}=(1+t^6)(1+t^3)(1-t)(1+t+t^2).\\ \therefore~L=\dfrac{1*(-1)}{2*2*1*3}= - \dfrac 1 {12}= - 0.833333.

Niranjan Khanderia - 3 years, 11 months ago

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