লিমিট
limx→0 x3cos(1x)=?\underset{x\rightarrow0}{lim}\ x^3\cos{\left(\frac{1}{x}\right)=?}x→0lim x3cos(x1)=?
0
1
(−1)\left(-1\right)(−1)
e
−1≤cos(1x)≤1⇒−x3≤x3cos(1x)≤x3-1\le\cos{\left(\frac{1}{x}\right)}\le1\Rightarrow-x^3\le x^3\cos{\left(\frac{1}{x}\right)}\le x^3−1≤cos(x1)≤1⇒−x3≤x3cos(x1)≤x3 ∴limx→0 x3cos(1x)=0\therefore \underset {x→0}{lim}\ x^3cos\left ( \frac{1}{x} \right )=0∴x→0lim x3cos(x1)=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
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limx→0 (1+5x)3x+2x= ?\mathrm{\lim\limits_{x\rightarrow0}\ \left(1+5x\right)^\frac{3x+2}{x}=\ ? }x→0lim (1+5x)x3x+2= ?