JEE-Mains 2014 (10/30)

Calculus Level 2

lim ⁡ x → 0 sin ⁡ ( π cos ⁡ 2 x ) x 2 = ? \large \displaystyle \lim_{x \to 0} \dfrac{\sin(\pi \cos^2x)}{x^2}= \, ?

− π -\pi 1 1 π 2 \frac{\pi}{2} π \pi

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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3 solutions

Sandeep Rathod
Jun 15, 2015

lim ⁡ x → 0 sin ⁡ ( π cos ⁡ 2 x ) x 2 = lim ⁡ x → 0 sin ⁡ ( π ( 1 − sin ⁡ 2 x ) x 2 \displaystyle \lim_{x \to 0} \dfrac{\sin(\pi \cos^2x)}{x^2}= \lim_{x \to 0} \dfrac{\sin(\pi(1 - \sin^2x)}{x^2}

lim ⁡ x → 0 sin ⁡ ( π sin ⁡ 2 x ) x 2 \lim_{x \to 0} \dfrac{\sin(\pi \sin^2x)}{x^2}

lim ⁡ x → 0 sin ⁡ ( π sin ⁡ 2 x ) π sin ⁡ 2 x × π sin ⁡ 2 x x 2 \lim_{x \to 0} \dfrac{\sin(\pi \sin^2x)}{\pi\sin^2x} \times \frac{\pi\sin^2x}{x^2}

= π = \pi

28 Helpful 4 Interesting 4 Brilliant 14 Confused

Moderator note:

For x ≈ 0 x \approx 0 , you can apply sin ⁡ ( π sin ⁡ 2 x ) ≈ π sin ⁡ 2 x \sin(\pi \sin^2 x) \approx \pi \sin^2 x .

We can use l'Hopital rule

Refaat M. Sayed - 5 years, 12 months ago

In case people don't get it, Sandeep is using the fact that

x ∼ s i n x x \sim sin x

at x = 0 x=0 .

Jake Lai - 5 years, 12 months ago

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what is this? is he an engineer?

Halim Amran - 11 months, 3 weeks ago

Nice solution but how did you obtain the third expression and where did the minus sign go??

Puneet Pinku - 5 years, 2 months ago

As for me, I think we can use l'hospital's rule.. But before doing that we can first apply the equivalence of squeezes theorem as follows.. X≈sinx which is similar to sin(πcos²x) ≈πcos²x Lim as x appro 0

Sin(πcos²x)/(x²) =Lim as x appro 0, πcos²x/x² now we can take the derivative as follows.. π (cos²x)'/(x²)' = (-2πsinx)/2x Lim as x appro 0, -π(1)= -π

Gerald Sohn - 1 year, 8 months ago
Muhammad Ardivan
Jun 17, 2015

Yeah it's true cause lim x-->0 sinx/x =1

2 Helpful 0 Interesting 0 Brilliant 0 Confused

How he made 1-sin^2x as sin^2x

Tarun B - 4 years, 3 months ago

using tha angle difference formula sin(a-b)=sin(a)cos(b)-sin(b)cos(a) thus we have sin(pi-pi sin^2x)=sin(pi)cos(pi sin^2 (x))-sin(pi sin^2(x))cos(pi) since cos(pi)=-1 and sin(pi)=0 we get sin(pi-pi sin^2x)=sin(pi*sin^2(x))

Federico Binaglia - 1 year ago

Use L'Hopital's rule!

1 Helpful 0 Interesting 0 Brilliant 0 Confused

If we multiply by x in the numerator and the denominator we would get an indeterminate form. We could applu L'Hopital's rule to that?

jose esteban beleño barrozo - 7 months ago

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