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Find the limit:

$$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

Given that $f(x) = \cos(x^2)$

Any help will be appreciated.

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    $\begingroup$ Are you allowed to recognize this as the derivative and use the chain rule, or are you supposed to do this the hard way, and basically rederive the chain rule and the derivative of $\cos$ at the same time? $\endgroup$ – Arthur Feb 14 '18 at 7:57
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    $\begingroup$ What have you attempted thus far? $\endgroup$ – Mr Pie Feb 14 '18 at 8:00
  • $\begingroup$ i think the chain rule will be acceptable $\endgroup$ – fds Feb 14 '18 at 8:04
  • $\begingroup$ Then did you try to use the derivative? $\endgroup$ – Yuriy S Feb 14 '18 at 8:06
  • $\begingroup$ i have seen some similar question which is the same expression but it switches f(x) = cos (x^2). Is it the same kind of calculation? $\endgroup$ – fds Feb 14 '18 at 8:07
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This is the definition of the derivative of $f(y)=\cos (y^2)$ at the point $y=x$. So, $$ \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=f'(x)=-2x\sin(x^2) $$

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  • $\begingroup$ Thank you for helping $\endgroup$ – fds Feb 14 '18 at 8:13
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The derivative can be found by first principles.

Method: use sum to product identity. $$ \begin{align*} \lim_{h \rightarrow0}\frac{\cos(x+h)^2-\cos x^2}{h} &= \lim_{h \rightarrow0}\frac{-2\sin(x^2+hx+h^2/2)\sin(hx+h^2/2)}{h} \\ &=\lim_{h \rightarrow0}-2\sin(x^2+hx+h^2/2) \frac{\sin(hx+h^2/2)}{hx+h^2/2}\frac{hx+h^2/2}{h} \\ &=\lim_{h \rightarrow0}-2\sin(x^2+hx+h^2/2) \frac{\sin(hx+h^2/2)}{hx+h^2/2}(x+h/2)\\ &=-2\sin(x^2)(1)(x)=-2x\sin(x^2) \end{align*} $$

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  • $\begingroup$ Thank you for helping $\endgroup$ – fds Feb 14 '18 at 8:13
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$$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

Let $f(x) = \cos(x^2)$, then

$$\lim_{h \to 0} \frac{\cos((x+h)^2)-\cos(x^2)}{h}$$ Apply L'Hopital's Rule $$\lim _{h\to 0}\left(\frac{-2\sin \left(\left(x+h\right)^2\right)\left(x+h\right)}{1}\right)$$ $$ \lim _{h\to 0}\left(-2\left(x+h\right)\sin \left(\left(x+h\right)^2\right)\right)$$ $$-2\left(x+0\right)\sin \left(\left(x+0\right)^2\right)$$

$$-2x\sin \left(x^2\right)$$



You can also apply the Chain rule: $f(g(x)) = (f\circ g)'=(f'\circ g)\cdot g'$, where in your case $f(x) = \cos(x)$ and $g(x) = x^2$

$$f(g(x)) = \cos(x^2) = -\sin(x^2)\cdot2x = -2x\sin(x^2)$$

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