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What is the value of the following limit? $$\lim_{x \to 0} \frac{d}{dx}\,\frac{\sin^2 x}{x}$$
A) $0$
B) $2$
C) $1$
D) $\frac{1}{2}$

Any quick or easy way to do this, without differentiating it first ?

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    How would you do it without differentiating? Do you want to expand the differential by definition, and try to exchange limits or something? – 2017-01-31
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    that's what exactly i asked, I have no idea how can we proceed without differentiation. I suspect if there is any such property, if its not then i will close this question for discussion. – 2017-01-31
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    No, unfortunately I cannot see any other method. I think you should close this question for discussion. – 2017-01-31
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    Its kind of like asking: compute $3+5$ without adding. – 2017-01-31
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    we can do this too..using counting :D http://www.preschoolmath.info/preschool-math-worksheeets/addition/pictures-addition/pictures-addition9.gif – 2017-02-01

4 Answers 4

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Personally, I'd say that differentiating is the quick and easy way to do it.

However, a way to guess the numerical value knowing beforehand that it exists is observing that, if you apply l'Hopital to $$\lim_{x\to 0}\frac{\sin^2 x}{x^2}=\lim_{x\to 0}\frac{\frac{\sin^2 x}{x}}{x}=\lim_{x\to 0}\frac{\frac{\sin x}{x}\cdot\sin x}{x}=\left[\frac{1\cdot 0}0\right]$$

you'll end up evaluating exactly that limit there.

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    How did this answer to my question ? It does not take differentiation into account, or it does ? – 2017-01-31
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    L'Hopital yields that, if $\lim\limits_{x\to 0} \frac d{dx}\frac{\sin^2 x}{x}$ exists, then it is equal to $\lim\limits_{x\to 0}\frac{\sin^2x}{x^2}$. And that should be all you need to know. – 2017-01-31
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    Not clear :( How it making sense ? – 2017-01-31
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    That's all there is to it. "Quick and easy" tricks are like that: you either know the fact they exploit or you do not. – 2017-01-31
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A possible solution using Taylor series.

$$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$$ $$\sin^2(x)=x^2-\frac{x^4}{3}+O\left(x^6\right)$$ $$\frac{\sin^2(x)}x=x-\frac{x^3}{3}+O\left(x^5\right)$$ $$\left(\frac{\sin^2(x)}x\right)'=1-x^2+O\left(x^3\right)$$

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We will first differentiate the expression, then take the limit.

Let $y=\frac{\sin^2 x}{x}$.

$$\therefore \ln{y}=\ln(\sin^2x)-\ln x$$

Using logarithmic differentiation:

$$y'=\left[\frac{x\sin2x-\sin^2x}{x\sin^2x}\right]\cdot\left[\frac{\sin^2x}{x}\right]$$

Simplifying:

$$y'=\frac{x\sin 2x-\sin^2x}{x^2}$$

Now taking the limit:

$$\lim_{x\to0}\frac{x\sin 2x-\sin^2x}{x^2}$$

L'Hopital

$$=\lim_{x\to0}\frac{\sin2x+2x\cos x-\sin2x}{2x}$$

Cancelling terms:

$$\lim_{x\to0}(\cos x)=1$$

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    The question clearly says dont differentiate – 2017-01-31
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    I guess that's why I'm a math major and not a literature major. – 2017-01-31
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    Just saying man. Read before you answer – 2017-01-31
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    That's fair. I'm leaving this here until someone does it without differentiating. – 2017-01-31
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Let $f(x) =(\sin^{2}x)/x,f(0)=0$ then we can see easily by definition of derivative that $f'(0)=1$. The limit in question is $\lim_{x\to 0}f'(x)$ and if it exists it has to be $f'(0)=1$. That the limit exists follows from the fact that $f'(x) $ is continuous. But I think you can't avoid the act of differentiation.