Question Number 99938 by bemath last updated on 24/Jun/20 $$\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\frac{\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}+\sqrt{\mathrm{x}}−\mathrm{1}}{\:\sqrt{\mathrm{x}−\mathrm{1}}}\:?\: \\ $$ Commented by bobhans last updated on 24/Jun/20 $$\underset{\mathrm{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\:\frac{\frac{\mathrm{x}}{\:\sqrt{\mathrm{x}^{\mathrm{2}}…
Question Number 34375 by rahul 19 last updated on 05/May/18 $$\boldsymbol{{T}}{he}\:{value}\:{of}\: \\ $$$${lim}_{{x}\rightarrow\frac{\pi}{\mathrm{2}}} \:\:\:\frac{\left[\frac{{x}}{\mathrm{2}}\right]}{\mathrm{log}\:\left(\mathrm{sin}\:{x}\right)}\:=\:? \\ $$$$\left[.\right]=\:{greatest}\:{integer}\:{function}. \\ $$ Answered by MJS last updated on 05/May/18…
Question Number 34374 by rahul 19 last updated on 05/May/18 $${lim}_{{x}\rightarrow\mathrm{1}} \left\{\mathrm{1}−{x}+\left[{x}−\mathrm{1}\right]+\left[\mathrm{1}−{x}\right]\right\}\:=\:? \\ $$$$\left[.\right]=\:{greatest}\:{integer}\:{function}. \\ $$ Answered by MJS last updated on 05/May/18 $$\left[{x}−\mathrm{1}\right]=\left[{x}\right]−\mathrm{1} \\…
Question Number 34367 by rahul 19 last updated on 05/May/18 $$\boldsymbol{\mathrm{Evaluate}}\: \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow\infty} \left(\frac{{n}}{{n}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }+\frac{{n}}{{n}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }+…..+\frac{{n}}{{n}^{\mathrm{2}} +{n}^{\mathrm{2}} }\right). \\ $$ Answered by tanmay.chaudhury50@gmail.com…
Question Number 99761 by Lekhraj last updated on 23/Jun/20 Commented by Dwaipayan Shikari last updated on 23/Jun/20 $${Ans}=\mathrm{1} \\ $$ Commented by Dwaipayan Shikari last…
Question Number 99720 by bobhans last updated on 23/Jun/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:−\:\frac{\mathrm{1}}{\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}\:? \\ $$ Commented by john santu last updated on 23/Jun/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:^{\mathrm{2}}…
Question Number 99713 by Ar Brandon last updated on 22/Jun/20 $$\mathrm{Given}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{nx}^{\mathrm{n}+\mathrm{1}} −\left(\mathrm{n}+\mathrm{1}\right)\mathrm{x}^{\mathrm{n}} +\mathrm{1}}{\mathrm{x}^{\mathrm{p}+\mathrm{1}} −\mathrm{x}^{\mathrm{p}} −\mathrm{x}+\mathrm{1}}\:,\:\mathrm{x}\in\mathbb{R}\:\:\mathrm{and}\:\:\left(\mathrm{n},\mathrm{p}\right)\in\mathbb{N}^{\ast} ×\mathbb{N}^{\ast} \\ $$$$\mathrm{a}\backslash\mathcal{C}\mathrm{alculate}\:\underset{\mathrm{x}\rightarrow+\infty} {\mathrm{lim}f}\left(\mathrm{x}\right) \\ $$$$\mathrm{b}\backslash\mathrm{Show}\:\mathrm{that}\:\underset{\mathrm{x}\rightarrow\mathrm{1}} {\mathrm{lim}}=\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2p}} \\ $$ Answered…
Question Number 34160 by candre last updated on 01/May/18 $${prove}\:{that} \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}ln}\:{x}\centerdot\mathrm{ln}\:\left(\mathrm{1}+{x}\right)=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}ln}\:{x}\centerdot\mathrm{ln}\:\left(\mathrm{1}+{x}\right) \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated on 01/May/18 Commented…
Question Number 99697 by Ar Brandon last updated on 22/Jun/20 $$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{2n}} {\sum}}\frac{\mathrm{k}}{\mathrm{k}+\mathrm{n}^{\mathrm{2}} } \\ $$ Commented by MWSuSon last updated on 22/Jun/20 just dropping a comment so that I'll get notified when someone solves it. if only the k in the denominator was k^2��…
Question Number 99685 by Ar Brandon last updated on 22/Jun/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{limits}\:\mathrm{when}\:\mathrm{n}\:\mathrm{goes}\:\mathrm{to}\:\mathrm{infinty}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{summation}\:\mathrm{series}; \\ $$$$\mathrm{a}\backslash\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{E}\left(\mathrm{kx}\right),\:\:\mathrm{x}\in\mathbb{R} \\ $$$$\mathrm{b}\backslash\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\begin{pmatrix}{\mathrm{n}}\\{\mathrm{k}}\end{pmatrix}^{−\mathrm{1}} \\ $$ Commented…