লিমিট
limx→π(1−4tanx)cotx=\displaystyle \lim_{\mathrm{x}\rightarrow \pi }(1- 4 \tan \mathrm{x} )^{\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}}=x→πlim(1−4tanx)cotx=
e\mathrm{e}e
e4\mathrm{e}^{4}e4
e−1\mathrm{e}^{-1}e−1
e−4\mathrm{e}^{-4}e−4
Let, L=limx→π(1−4tanx)cotx\text{L}= \displaystyle \lim_{\mathrm{x}\rightarrow \pi }(1- 4 \tan \mathrm{x} )^{\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}}L=x→πlim(1−4tanx)cotx
Put y=π−x⇒x→π⇔y→0y=\pi-x\Rightarrow x\to \pi\Leftrightarrow y\to 0y=π−x⇒x→π⇔y→0
∴L=limy→0(1−4tan(π−y))cot(π−y)\therefore \text{L}= \displaystyle \lim_{\mathrm{y}\rightarrow 0 }(1- 4 \tan \mathrm{(\pi-y)} )^{\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{(\pi-y)}}∴L=y→0lim(1−4tan(π−y))cot(π−y)
=limy→0(1+4tany)−coty[∵tan(π−y)=−tany]\quad \quad = \displaystyle \lim_{\mathrm{y}\rightarrow 0 }(1+4 \tan \mathrm{y} )^{-\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{y}}[\because \tan (\pi-y)=-\tan y]=y→0lim(1+4tany)−coty[∵tan(π−y)=−tany]
Clearly form of the limit is 1∞1^{\infty}1∞
∴L=limy→0(1+4tany)14tany×(−4)=e−4\therefore \text{L}= \displaystyle \lim_{\mathrm{y}\rightarrow 0 }(1+4 \tan \mathrm{y} )^{\dfrac{1}{4\tan y}\times(-4)}=e^{-4}∴L=y→0lim(1+4tany)4tany1×(−4)=e−4
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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0. limx→0(1+5x)13x \lim_{x \to 0} \left ( 1 + 5 x \right )^{\frac{1}{3 x}} limx→0(1+5x)3x1 এর মান নিচের কোনটি ?
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