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How to find the value of this limit by using the definiton of deriviatve :

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

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

2 Answers

up vote 2 down vote accepted

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.

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

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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