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limx→0sinx−loge(excosx)xsinx \lim_{x → 0} \frac{\sin{x} - \log_{e}{\left ( e^{x} \cos{x} \right )}}{x \sin{x}} limx→0xsinxsinx−loge(excosx) এর মান কোনটি?
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limx→0sinx−loge(excosx)xsinx=limx→0sinx−x−loge(cosx)xsinx \lim _{x \rightarrow 0} \frac{\sin x-\log _{e}\left(e^{x} \cos x\right)}{x \sin x}=\lim _{x \rightarrow 0} \frac{\sin x-x-\log _{e}(\cos x)}{x \sin x} limx→0xsinxsinx−loge(excosx)=limx→0xsinxsinx−x−loge(cosx)
Using L' Hospital' s law, limx→0cosx−1+tanxsinx+xcosx;[00]=limx→0−sinx+sec2xcosx+cosx−xsinx=11+1=12 \lim _{x \rightarrow 0} \frac{\cos x-1+\tan x}{\sin x+x \cos x} ;\left[\frac{0}{0}\right]=\lim _{x \rightarrow 0} \frac{-\sin x+\sec ^{2} x}{\cos x+\cos x-x \sin x}=\frac{1}{1+1}=\frac{1}{2} limx→0sinx+xcosxcosx−1+tanx;[00]=limx→0cosx+cosx−xsinx−sinx+sec2x=1+11=21
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
