মিশ্র ফাংশন সংক্রান্ত
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e-θ
limθ → 0 θsin θ = lim θ →0 1sinθθ= 11= 1\lim_{\theta\ \rightarrow\ 0}\ \frac{\theta}{\sin\ \theta}\ =\ \lim\ _{\theta\ \rightarrow0}\ \frac{1}{\frac{\sin\theta}{\theta}}=\ \frac{1}{1}=\ 1limθ → 0 sin θθ = lim θ →0 θsinθ1= 11= 1
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.
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
limx→02xsinb2x=? \lim_{x → 0} 2^{x} \sin \frac{b}{2^{x}} = ? limx→02xsin2xb=?