1
$\begingroup$

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

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

$\endgroup$
4
  • 3
    $\begingroup$ Shouldn't it be $\sin{\left((x+h)^{2}\right)}$? $\endgroup$ – Thomas Russell Aug 5 '12 at 0:25
  • 1
    $\begingroup$ $f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}$. 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
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.