TIFR CSE 2011 | Part A | Question: 14

Question

The limit $$\lim_{x \to 0} \frac{d}{dx}\,\frac{\sin^2 x}{x}$$ is

• A. $0$
• B. $2$
• C. $1$
• D. $\frac{1}{2}$
• E. None of the above

Answer 1

Answer is 1.

$\qquad\lim\limits_{x\to 0} \dfrac{d}{dx} \dfrac{sin^{2}x}{x}$

 

Find $\;\dfrac{d}{dx} \left(\dfrac{\sin^{2}x}{x}\right)$

$=\dfrac{(\sin^{2}x)^{1}x-(x)^{1}\sin^{2}x}{x^{2}}$

$=\dfrac{\left[2\sin x.\cos x\right]x-sin^{2}x}{x^{2}}$

$=\dfrac{x\sin {2x}-\sin^{2}x}{x^{2}}$

$\lim\limits_{x\to 0}\left(\dfrac{x \sin {2x}- \sin^{2}x}{x^2}\right)$

$=\lim\limits_{x\to 0}\left(\dfrac{x \sin {2x}}{x^2}- \dfrac{\sin^{2}x}{x^{2}}\right)$

$=\lim\limits_{x\to 0}\dfrac{x \sin {2x}}{x^2}- \lim\limits_{x\to 0} \left(\dfrac{\sin x}{x}\right)^{2}$

$=\lim\limits_{x\to 0}\dfrac{\cos {2x}.2}{1}- 1$

$=2 [\cos 2(0)] - 1$

$=2(1) - 1\Rightarrow 1.$

Comments

hey .. i have a doubt ? i the above question why cant we apply L'hospital rule and under this rules we differntiate denominator n numerator seperately and then apply limit if not in intermediate form..and then apply limit directly ??? is i am right ?

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Here given d/dx(sin^2x/x ), so first we have to find inner most value and apply limit.

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ohh.. my mistake this for pointing out

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i did same mistake

Answer 2

As we know $$\frac{lim}{x \to 0} \frac{sinx}{x}=1$$

 

$\large \frac{lim}{x \to 0}\ \frac{\partial }{\partial x}\frac{sin^2x}{x}\\ \frac{lim}{x \to 0}\ \frac{\partial }{\partial x}\left ( \frac{sinx}{x} \right ) sinx\\ \frac{lim}{x \to 0}\ \frac{\partial }{\partial x}sinx\\ \frac{lim}{x \to 0}\ cosx\\=1$

Comments

Although you got the right answer. But are you sure you can apply limit before solving differentiation?

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is it right approach?

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@bhuv Can you give a link to any other example where we apply limit before solving? Or can you give reasoning as to why we can do this?

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This is like

121/11 

= 121/11  (because 2 = 1+1 so cancel cancel)

= 11

(Unless we can actually apply limit like that, then I am sorry.)

Answer 3

1-2sin^2(x)=cos(2x)
  d/dx ((1-cos(2x))/2x) = {2x(2sin(2x) - (1-cos2x)(2)}/4x^2

   = 4sin(2x)/(4x) - (4sin^2 (x)/ 4x^2)

=  2 sin(2x)/2x - sin^2(x)/x^2

Lt x->0 sin x/x =1

Lt x->0 (2 sin(2x)/2x - sin^2(x)/x^2)

= (2 Lt x->0 sin(2x)/2x - Lt x->0(sin x/x) . Lt x->0 (sin x/ x) )

= 2 - 1

= 1

Answer 4

It is given
$$\lim_{x \to 0} \frac{d}{dx}\,\frac{\sin^2 x}{x} = \lim_{x \to 0} \frac{d}{dx} \sin^2(x)x^{-1}$$

Using chain rule, we get
$$\lim_{x \to 0} [-1.\sin^2(x)x^{-2} + \sin(2x)x^{-1}]$$

$$-1.\Big(\underbrace{\lim_{x \to 0} \frac{\sin(x)}{x}}_{1}\Big).\Big(\underbrace{\lim_{x \to 0} \frac{\sin(x)}{x}}_{1}\Big) + 2.\Big(\underbrace{\lim_{ x \to 0} \frac{\sin(2x)}{2x}}_{1}\Big)$$ 

This gives, $-1.1.1 + 2.1 = 2-1=1$

$\textbf{Option (C) is correct}$