মিশ্র ফাংশন সংক্রান্ত
limx→01−cos2xx\underset{x \rightarrow 0}{lim}\frac{\sqrt{1 - c o s 2 x}}{x}x→0limx1−cos2x এর মান কত?
222\sqrt{2}22
2\sqrt{2}2
12\frac{1}{\sqrt{2}}21
limx→01−cos2xxlimx→02sin2xxlimx→02sinxx=2limx→0sinxx=2⋅1=2 \begin{aligned} & \lim _{x \rightarrow 0} \frac{\sqrt{1-\cos 2 x}}{x} \\ & \lim _{x \rightarrow 0} \frac{\sqrt{2 \sin ^{2} x}}{x} \\ & \lim _{x \rightarrow 0} {\sqrt{2}}{\frac{\sin x}{x}} \\ = & \sqrt{2} \lim _{x \rightarrow 0} \frac{\sin x}{x} \\ = & \sqrt{2} \cdot 1 \\ = & \sqrt{2}\end{aligned} ===x→0limx1−cos2xx→0limx2sin2xx→0lim2xsinx2x→0limxsinx2⋅12
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.
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
limx→02xsinb2x=? \lim_{x → 0} 2^{x} \sin \frac{b}{2^{x}} = ? limx→02xsin2xb=?
