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Limx→ax52−a52x−a\underset {x→a} {Lim} \frac{x^\frac{5}{2}-a^\frac{5}{2}}{\sqrt x-\sqrt a}x→aLimx−ax25−a25 এর মান কত?
5x25x^25x2
−5x2-5x^2−5x2
5a25a^2 5a2
−5a2-5a^2−5a2
ধরি, x=y \sqrt x=y x=y ও a=b ∴x→a\sqrt a=b\ \ \therefore x\rightarrow a a=b ∴x→a হলে, y→by\rightarrow by→b
∴Limx→ax52−a52x−a =Limy→by5−b5y−b=5b5−1=5b4\therefore \underset {x→a} {Lim} \frac{x^\frac{5}{2}-a^\frac{5}{2}}{\sqrt x-\sqrt a}\ = \underset {y→b} {Lim} \frac{y^5-b^5}{y-b}=5b^{5-1}=5b^4∴x→aLimx−ax25−a25 =y→bLimy−by5−b5=5b5−1=5b4
=5(a)4=5a2 =5\left(\sqrt a\right)^4=5a^2\ =5(a)4=5a2 [∴Limx→axn−anx−a=nan−1 ]\left[\therefore \underset {x→a}{Lim} \frac{x^n-a^n}{x-a}=na^{n-1}\ \right] [∴x→aLimx−axn−an=nan−1 ]
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?
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= ?
এর সঠিক মান কোনটি?