লিমিট
limx→0(sinx−xx)(sin1x)\displaystyle \lim_{x\rightarrow 0}(\frac{\sin x-x}{x})(\sin\frac{1}{x})x→0lim(xsinx−x)(sinx1) is:
does not exist
is equal to 0
is equal to 1
exists and different from 0 and 1
sinx=x−x33!+x55!....(1)sin x = x -\dfrac{x^{3}}{3!}+\dfrac{x^{5}}{5!}....(1)sinx=x−3!x3+5!x5....(1)
sin x−x=−x33!+x55!−x77!sin\ x-x =-\dfrac{x^{3}}{3!}+\dfrac{x^{5}}{5!}-\dfrac{x^{7}}{7!}sin x−x=−3!x3+5!x5−7!x7
sin x−xx=−x23!+x45!−x67!....=0 (as x found to 0)\dfrac{sin\ x-x}{x} =-\dfrac{x^{2}}{3!}+\dfrac{x^{4}}{5!}-\dfrac{x^{6}}{7!}....=0\ (as\ x\ found\ to\ 0)xsin x−x=−3!x2+5!x4−7!x6....=0 (as x found to 0)
sin(1x)(\dfrac{1}{x})(x1) is an oscillatory function means same finite no.
0×finite no.=00\times finite\ no.=00×finite no.=0
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
এর সঠিক মান কোনটি?
0. limx→0(1+5x)13x \lim_{x \to 0} \left ( 1 + 5 x \right )^{\frac{1}{3 x}} limx→0(1+5x)3x1 এর মান নিচের কোনটি ?
এর সঠিক মান কোনটি?