Question

$\lim_{h \to \0} (cos(x+h)-cosx)/h$



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Answer 1

Given,

$\displaystyle\longrightarrow\sf{\lim_{h\to0}\dfrac {\cos (x+h)-\cos x}{h}}$

We know the sum - to-product rule,

$\displaystyle\longrightarrow\sf{\cos A-\cos B=-2\sin\left (\dfrac {A+B}{2}\right)\cos\left (\dfrac {A-B}{2}\right)}$

Then,

$\displaystyle\longrightarrow\sf{\lim_{h\to0}\dfrac {\cos (x+h)-\cos x}{h}=\lim_{h\to0}\dfrac {-2\sin\left (\dfrac {x+h+x}{2}\right)\sin\left (\dfrac {x+h-x}{2}\right)}{h}}$

$\displaystyle\longrightarrow\sf{\lim_{h\to0}\dfrac {\cos (x+h)-\cos x}{h}=-\lim_{h\to0}\sin\left (x+\dfrac {h}{2}\right)\cdot2\lim_{h\to0}\dfrac {\sin\left (\dfrac {h}{2}\right)}{h}}$

Well, as $\sf{h\longrightarrow0,\ \dfrac {h}{2}\longrightarrow0.}$ Then,

$\displaystyle\longrightarrow\sf{\lim_{h\to0}\dfrac {\cos (x+h)-\cos x}{h}=-\sin\left (x+\dfrac {0}{2}\right)\cdot\lim_{\frac{h}{2}\to0}\dfrac {\sin\left (\dfrac {h}{2}\right)}{\left (\dfrac {h}{2}\right)}}$

$\displaystyle\longrightarrow\sf{\lim_{h\to0}\dfrac {\cos (x+h)-\cos x}{h}=-\sin x\cdot1}$

$\displaystyle\longrightarrow\sf{\underline {\underline {\lim_{h\to0}\dfrac {\cos (x+h)-\cos x}{h}=-\sin x}}}$

Answer 2

Answer:

$\displaystyle \red{{-\sin x}}$

Step-by-step explanation:

Given :

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

Using formula cos c - cos d = - sin [ ( c + d ) / 2 ] sin [ ( c - d ) / 2 ]

$\displaystyle{ \lim_{h \to 0} -2\left(\frac{\sin \left(\dfrac{2x+h)}{2}\right)\times \sin \left( \dfrac{ h}{2} \right)}{h} \right)}$

Now multiply and divide by 1 / 2 :

$\displaystyle{ \lim_{h \to 0} -\frac{2}{2} \left(\frac{\sin \left(\dfrac{2x+h)}{2}\right)\times \sin \left( \dfrac{ h}{2} \right)}{\dfrac{h}{2}}\right)}$

We know :

$\displaystyle{ \lim_{x \to 0} \frac{\sin x}{x} = 1}\\\\\\\displaystyle{\implies \lim_{h/2 \to 0} \frac{\sin h/2}{h/2} = 1}$

# As h approaches to 0 that implies h / 2 also approaches to 0.

$\displaystyle{ \lim_{h \to 0} -\sin \left(\dfrac{2x+h}{2}\right)}$

Now putting h = 0

$\displaystyle{ \lim_{h \to 0} -\sin \left(\dfrac{2x+0}{2}\right)}\\\\\\\displaystyle{\rightarrow -\sin \left(\dfrac{2x}{2}\right)}\\\\\\\displaystyle \boxed{{-\sin x}}$

Therefore we get answer.