লিমিট
Ltx→0(1+kx)1x\operatorname{Lt}_{\mathrm{x} \rightarrow 0}(1+\mathrm{kx})^{\frac{1}{\mathrm{x}}}Ltx→0(1+kx)x1
Ltx→0(1+kx)1x=Ltx→0(1+kx)1kx⋅k=Ltx→0{(1+kx)1kx}k=ek \operatorname{Lt}_{\mathrm{x} \rightarrow 0}(1+\mathrm{kx})^{\frac{1}{\mathrm{x}}}=\underset{\mathrm{x} \rightarrow 0}{\operatorname{Lt}}(1+\mathrm{kx})^{\frac{1}{\mathrm{kx}} \cdot \mathrm{k}}=\operatorname{Lt}_{\mathrm{x} \rightarrow 0}\left\{(1+\mathrm{kx})^{\frac{1}{\mathrm{kx}}}\right\}^{\mathrm{k}}=\mathrm{e}^{\mathrm{k}} Ltx→0(1+kx)x1=x→0Lt(1+kx)kx1⋅k=Ltx→0{(1+kx)kx1}k=ek
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
