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4 Answers

Best answer
17 votes
17 votes
Answer is 1.

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

 

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

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

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

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

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

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

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

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

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

$=2(1) - 1\Rightarrow 1.$
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4 votes
4 votes
As we know $$\frac{lim}{x \to 0} \frac{sinx}{x}=1$$

 

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

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

0 votes
0 votes

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


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

Answer:

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