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limx→−1cosx−cos2xx2−∣x∣\lim _ { x \rightarrow - 1 } \dfrac { \cos x - \cos 2 x } { x ^ { 2 } - | x | }limx→−1x2−∣x∣cosx−cos2x is equal to?
-∞∞ ∞
cos2\cos 2cos2
2sin22 \sin 22sin2
None
limx→−1cosx−cos2xx2−∣x∣=limx→−1cos(x)−cos2xx2+x=cos(−1)−cos(−2)1−1=cos(1)−cos(2)0=−∞[∵cos1<cos2] \begin{aligned} & \lim _{x \rightarrow-1} \frac{\cos x-\cos 2 x}{x^{2}-|x|} \\ = & \lim _{x \rightarrow-1} \frac{\cos (x)-\cos 2 x}{x^{2}+x} \\ = & \frac{\cos (-1)-\cos (-2)}{1-1}=\frac{\cos (1)-\cos (2)}{0}=-\infty[\because \cos 1<\cos 2]\end{aligned} ==x→−1limx2−∣x∣cosx−cos2xx→−1limx2+xcos(x)−cos2x1−1cos(−1)−cos(−2)=0cos(1)−cos(2)=−∞[∵cos1<cos2]
limx→0+(cosecx)1/logx\displaystyle \lim_{x\rightarrow 0^{+}}{(\cosec x)^{1/\log x}}x→0+lim(cosecx)1/logx=?
If f′f 'f′ (0) = 0 and f(x) is a differentiable and increasing function,then lim x→0 x \rightarrow 0x→0 x.f′(x2)f′(x)\frac {x.f ' (x^2)}{f ' (x)}f′(x)x.f′(x2)
ddx(9x)= \frac{d}{d x}\left(9^{x}\right)= dxd(9x)= কত?
The value of limx→−1π−cos−1xx+1\lim_{x \rightarrow -1} \dfrac{\sqrt{\pi}-\sqrt{\cos^{-1}x}}{\sqrt{x+1}}limx→−1x+1π−cos−1x is given by
