লিমিট
limx→0sin9x−sin7xsin7x−sin5x= ?\mathrm{\lim\limits_{x\rightarrow0}\frac{\sin{9x}-\sin{7x}}{\sin{7x}-\sin{5x}}=\ ? }x→0limsin7x−sin5xsin9x−sin7x= ?
1
−1-1 −1
2
−2-2−2
limx→02cos8xsinx2cos6xsinx=limx→0cos8xcos6x=1\mathrm{\lim\limits_{x\rightarrow0}\frac{2\cos{8x}\sin{x}}{2\cos{6x}\sin{x}}=\lim\limits_{x\rightarrow0}\frac{\cos{8x}}{\cos{6x}}=1 }x→0lim2cos6xsinx2cos8xsinx=x→0limcos6xcos8x=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.
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
