How to find the value of this limit by using the definition of derivative :

$$ \lim_{h\to 0}\frac{\sin (x^{2}+h)-\sin x^{2}}{h} $$

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    $\begingroup$ Shouldn't it be $\sin{\left((x+h)^{2}\right)}$? $\endgroup$ – Thomas Russell Aug 5 '12 at 0:25
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    $\begingroup$ $f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}$. From this you can get $\lim_{h\to 0}\frac{\sin (x^{2}+h)-\sin x^{2}}{h} = \sin' x^2 = \cos x^2$. $\endgroup$ – copper.hat Aug 5 '12 at 0:35
  • $\begingroup$ @copper.hat Won't it become $2x\cos{(x^2)}$? $\endgroup$ – ladaghini Aug 5 '12 at 0:47
  • $\begingroup$ @ladaghini: No; it would have if it was $\sin((x+h)^2)$, though. Notice the subtle difference. $\endgroup$ – Clive Newstead Aug 5 '12 at 0:58

First substitute $y=x^2$, so that you have

$$\lim_{h \to 0} \frac{\sin(y+h) - \sin y}{h}$$

Does this look familiar? It's the definition of the derivative of $\sin$ evaluated at the point $y$. Now substitute back.

Edit: Didn't see this had been answered by copper.hat in the comments.


$$\lim_{h\to 0}\frac{\sin(x^2+h)-\sin x^2} h$$ $$\lim_{h\to 0}\frac{\sin(x^2)\cos(h)+\sin(h)\cos(x^2)-\sin x^2} h$$ $$\lim_{h\to 0}\frac{\sin(x^2)(\cos(h)-1)} h +\lim_{h\to 0}\frac{\sin(h)\cos(x^2)} h$$

You can prove geometrically (see here, for example) that $\lim_{h\to 0}\frac{\sin h} h=1$, and that $\lim_{h\to 0}\frac{1-\cos h} h=0$ (see this question), which gives the appropriate limit.


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