Calculus 1- Find directly the derivative of a function f.

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

Thus

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

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

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

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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 $x=\pi/4$.