mathop {lim }limits_{x to 0} frac{{cos ax - c | vedclass

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Question

(B) ત્રિકોણમિતીય નિત્યસમ $\cos C - \cos D = 2 \sin \left( \frac{C+D}{2} \right) \sin \left( \frac{D-C}{2} \right)$ નો ઉપયોગ કરતા:$\mathop {\lim }\limi

Vedclass · Accepted Answer

$\frac{{{b^2} - {a^2}}}{2}$

Vedclass · Answer

(B) ત્રિકોણમિતીય નિત્યસમ $\cos C - \cos D = 2 \sin \left( \frac{C+D}{2} ight) \sin \left( \frac{D-C}{2} ight)$ નો ઉપયોગ કરતા:$\mathop {\lim }\limits_{x 	o 0} \frac{{\cos ax - \cos bx}}{{{x^2}}} = \mathop {\lim }\limits_{x 	o 0} \frac{{2 \sin \left( \frac{ax+bx}{2} ight) \sin \left( \frac{bx-ax}{2} ight)}}{{{x^2}}}$$= 2 \mathop {\lim }\limits_{x 	o 0} \left( \frac{\sin \left( \frac{a+b}{2}x ight)}{x} ight) \left( \frac{\sin \left( \frac{b-a}{2}x ight)}{x} ight)$$= 2 \cdot \left( \frac{a+b}{2} ight) \cdot \left( \frac{b-a}{2} ight) = 2 \cdot \frac{b^2 - a^2}{4} = \frac{b^2 - a^2}{2}$વૈકલ્પિક રીતે,$L$-Hospital ના નિયમનો ઉપયોગ કરતા:$\mathop {\lim }\limits_{x 	o 0} \frac{{\cos ax - \cos bx}}{{{x^2}}} = \mathop {\lim }\limits_{x 	o 0} \frac{{-a \sin ax + b \sin bx}}{{2x}}$$= \mathop {\lim }\limits_{x 	o 0} \frac{{-a^2 \cos ax + b^2 \cos bx}}{2} = \frac{-a^2(1) + b^2(1)}{2} = \frac{b^2 - a^2}{2}$

$\mathop {\lim }\limits_{x \to 0} \frac{{\cos ax - \cos bx}}{{{x^2}}} = $

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જો $\alpha, \beta$ એ $x^{2}+bx+c=0$ ના ભિન્ન બીજ હોય,તો $\lim _{x \rightarrow \beta} \frac{e^{2(x^{2}+bx+c)}-1-2(x^{2}+bx+c)}{(x-\beta)^{2}}$ ની કિંમત શોધો.

જો $f(a)=2, f^{\prime}(a)=1, g(a)=-1, g^{\prime}(a)=2$ હોય,તો જ્યારે $x$ એ $a$ ની નજીક જાય,ત્યારે $\frac{g(x) f(a)-g(a) f(x)}{x-a}$ ની લક્ષ કિંમત શોધો.

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