lim _{x rightarrow frac{pi}{2}}left(frac{1}{l | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) ધારો કે $L = \lim _{x \rightarrow \frac{\pi}{2}} \frac{\int_{x^3}^{(\pi/2)^3} \cos(t^{1/3}) dt}{(x-\pi/2)^2}$.અહીં લક્ષ $\frac{0}{0}$ સ્વરૂપમાં હો

Vedclass · Accepted Answer

$\frac{3 \pi^2}{8}$

Vedclass · Answer

(A) ધારો કે $L = \lim _{x \rightarrow \frac{\pi}{2}} \frac{\int_{x^3}^{(\pi/2)^3} \cos(t^{1/3}) dt}{(x-\pi/2)^2}$.અહીં લક્ષ $\frac{0}{0}$ સ્વરૂપમાં હોવાથી,આપણે $L$'$H$ôpital ના નિયમ અને Leibniz ના સંકલન નિયમનો ઉપયોગ કરીશું:$L = \lim _{x \rightarrow \frac{\pi}{2}} \frac{\frac{d}{dx} \int_{x^3}^{(\pi/2)^3} \cos(t^{1/3}) dt}{2(x-\pi/2)}$$L = \lim _{x \rightarrow \frac{\pi}{2}} \frac{0 - \cos((x^3)^{1/3}) \cdot \frac{d}{dx}(x^3)}{2(x-\pi/2)}$$L = \lim _{x \rightarrow \frac{\pi}{2}} \frac{-\cos(x) \cdot 3x^2}{2(x-\pi/2)}$ધારો કે $h = x - \frac{\pi}{2}$,તેથી $x = h + \frac{\pi}{2}$. જ્યારે $x \rightarrow \frac{\pi}{2}$,ત્યારે $h \rightarrow 0$.$L = \lim _{h \rightarrow 0} \frac{-\cos(h + \pi/2) \cdot 3(h + \pi/2)^2}{2h}$$L = \lim _{h \rightarrow 0} \frac{\sin(h) \cdot 3(h + \pi/2)^2}{2h}$કારણ કે $\lim _{h \rightarrow 0} \frac{\sin h}{h} = 1$,તેથી:$L = 1 \cdot \frac{3(\pi/2)^2}{2} = \frac{3 \cdot \pi^2/4}{2} = \frac{3 \pi^2}{8}$.

$\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{1}{\left(x-\frac{\pi}{2}\right)^2} \int_{x^3}^{\left(\frac{\pi}{2}\right)^3} \cos \left(t^{1/3}\right) d t\right)$ ની કિંમત શોધો.

Similar Questions

$\lim _{x \rightarrow 0} \frac{\cos 2x - \cos 3x}{\cos 4x - \cos 5x} = $

$\mathop {\lim }\limits_{x \to 0} {(\cos x)^{\cot x}}$ ની કિંમત શું છે?

$\mathop {\lim }\limits_{\alpha \to \pi /4} \frac{{\sin \alpha - \cos \alpha }}{{\alpha - \frac{\pi }{4}}} = $

જો $l_1 = \lim_{x \rightarrow 2^{+}} (x + [x])$,$l_2 = \lim_{x \rightarrow 2^{-}} (2x - [x])$ અને $l_3 = \lim_{x \rightarrow \pi/2} \frac{\cos x}{x - \pi/2}$ હોય,તો:

જો $f(a) = 2, f'(a) = 1, g(a) = -1, g'(a) = 2$ હોય,તો $\lim_{x \to a} \frac{g(x)f(a) - g(a)f(x)}{x - a}$ ની કિંમત શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module