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The following limit represents the derivative of a function $f$ at a point $a$. Evaluate the limit.

$$\lim\limits_{h \to 0 } \frac{\sin^2\left(\frac\pi 4+h \right)-\frac 1 2} h$$

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L'Hospital? ${}$ – Henning Makholm Oct 23 '12 at 0:22
Looks like he is supposed to use the limit definition of a derivative. I would conjecture that L'Hospital is a bit too advanced if he's intended to use the limit definition. – Joe Oct 23 '12 at 0:23

Write the limit as

$$\mathop {\lim }\limits_{h \to 0} \frac{{2{{\sin }^2}\left( {\frac{\pi }{4} + h} \right) - 1}}{{2h}}$$

Now use $$2{\sin ^2}x - 1 = - \cos \left( {2x} \right)$$


$$=\mathop {\lim }\limits_{h \to 0} \frac{{-\cos \left( {\frac{\pi }{2} + 2h} \right)}}{{2h}}$$

$$ = \mathop {\lim }\limits_{h \to 0} \frac{{\sin \left( {2h} \right)}}{{2h}} = 1$$

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Let $f(x)=\sin^2x$. We have, $f^{\prime}(x)=2\sin x\cos x$. In the other hand \begin{equation} \begin{array}{lll} \lim_{h\rightarrow 0}\frac{\sin^2\left(\frac{\pi}{4}+h\right)-\frac{1}{2}}{h}&=&\lim_{h\rightarrow 0}\frac{f\left(\frac{\pi}{4}+h\right)-f\left(\frac{\pi}{4}\right)}{h}\\ &=&f^{\prime}(\pi/2)\\ &=&2\sin(\pi/4)\cos(\pi/4)\\ &=&2(1/\sqrt{2})(1/\sqrt{2})=1. \end{array} \end{equation}

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I interpret the hint as asking you to identify the function and where the derivative is taken. There are a couple of natural choices.

If you let $f(x)=\sin^2(\pi/4+x)$, your limit expression, directly from the definition of derivative, is $f'(0)$. Now calculate $f'(x)$ using the ordinary rules of differentiation, and find $f'(0)$.

Or else let $f(x)=\sin^2 x$. Then our limit is the derivative of $f(x)$ at $x=\pi/4$.

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I think he is trying to find the derivative using the limit. – Navin Oct 23 '12 at 0:40
@Navin: It is possible. My read of the question goes the other way. If we take the interpretation that you make, there is a good answer by Peter Tamaroff. – André Nicolas Oct 23 '12 at 0:43

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