লিমিট
নিচের সীমার মান কোনটি? limθ→0cot(π2−θ)−cos(π2−θ)θ2\lim _{\theta \rightarrow 0} \frac{\cot \left(\frac{\pi}{2}-\theta\right)-\cos \left(\frac{\pi}{2}-\theta\right)}{\theta^{2}}limθ→0θ2cot(2π−θ)−cos(2π−θ)
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limθ→0cot(π2−θ)−cos(π2−θ)θ2=limθ→0tanθ−sinθθ2=limθ→0tanθ(1−cosθ)θ2=limθ→0tanθθ×limθ→02sin2θ2θ \begin{array}{l} \lim _{\theta \rightarrow 0} \frac{\cot \left(\frac{\pi}{2}-\theta\right)-\cos \left(\frac{\pi}{2}-\theta\right)}{\theta^{2}} \\ =\lim _{\theta \rightarrow 0} \frac{\tan \theta-\sin \theta}{\theta^{2}} \\ =\lim _{\theta \rightarrow 0} \frac{\tan \theta(1-\cos \theta)}{\theta^{2}} \\ =\lim _{\theta \rightarrow 0} \frac{\tan \theta}{\theta} \times \lim _{\theta \rightarrow 0} \frac{2 \sin ^{2} \frac{\theta}{2}}{\theta} \end{array} limθ→0θ2cot(2π−θ)−cos(2π−θ)=limθ→0θ2tanθ−sinθ=limθ→0θ2tanθ(1−cosθ)=limθ→0θtanθ×limθ→0θ2sin22θ
=1×limθ→0(sinθ2θ2)22×14×limθ→0θ=1×1×12×0=0 (Ans.) \begin{array}{l} =1 \times \lim _{\theta \rightarrow 0}\left(\frac{\sin \frac{\theta}{2}}{\frac{\theta}{2}}\right)^{2} 2 \times \frac{1}{4} \times \lim _{\theta \rightarrow 0} \theta \\ =1 \times 1 \times \frac{1}{2} \times 0=0 \text { (Ans.) } \end{array} =1×limθ→0(2θsin2θ)22×41×limθ→0θ=1×1×21×0=0 (Ans.)
Evaluate the following limits.
limx→02−x−2+xx\displaystyle\lim_{x\rightarrow 0}\dfrac{\sqrt{2-x}-\sqrt{2+x}}{x}x→0limx2−x−2+x.
limx→1(xx−1−1logx) \lim_{x \rightarrow 1} \left ( \frac{x}{x - 1} - \frac{1}{\log{x}} \right ) limx→1(x−1x−logx1) এর মান কত ?
If the function f(x)=(1−x)tanπx2f(x) = (1 - x)\tan \dfrac{{\pi x}}{2}f(x)=(1−x)tan2πx is continuous at x=1x = 1x=1 ,then f(1)=f(1)=f(1)=
limx→0xtan2x−2xtanx(1−cos2x)2 \displaystyle \lim _{x \rightarrow 0} \dfrac{x \tan 2 x-2 x \tan x}{(1-\cos 2 x)^{2}} x→0lim(1−cos2x)2xtan2x−2xtanx is equal to
