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limx→0sin2x+3x2x+sin3x\displaystyle \lim_{x\rightarrow 0} \dfrac {\sin 2x + 3x}{2x + \sin 3x}x→0lim2x+sin3xsin2x+3x is equal to
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15\dfrac {1}{5}51
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Does not exist
limx→0sin2x+3x2x+sin3xlimx→02cos2x+32+3cos3x [L’Hopital rule]⇒2+32+3⇒55=1 \begin{array}{l}\lim _{x \rightarrow 0} \frac{\sin 2 x+3 x}{2 x+\sin 3 x} \\ \lim _{x \rightarrow 0} \frac{2 \cos 2 x+3}{2+3 \cos 3 x} \text { [L'Hopital rule]} \\ \Rightarrow \frac{2+3}{2+3} \Rightarrow \frac{5}{5}=1\end{array} limx→02x+sin3xsin2x+3xlimx→02+3cos3x2cos2x+3 [L’Hopital rule]⇒2+32+3⇒55=1
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
