Question Number 184329 by greougoury555 last updated on 05/Jan/23 $$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}^{−\mathrm{1}} \left({p}+{x}\right)−\mathrm{tan}^{−\mathrm{1}} \left({p}−{x}\right)}{{x}}=? \\ $$ Answered by SEKRET last updated on 05/Jan/23 $$\:\:\frac{\mathrm{2}}{\mathrm{1}+\boldsymbol{\mathrm{p}}^{\mathrm{2}} } \\…
Question Number 184239 by mathlove last updated on 04/Jan/23 $${f}\left({x}\right)=\begin{cases}{{x}−\mathrm{2}\:\:\:\:\:\:\:{if}\:\:{x}>\mathrm{3}}\\{\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{if}\:\:{x}=\mathrm{3}}\\{−{x}+\mathrm{4}\:\:\:\:{if}\:\:\mathrm{1}\leqslant{x}<\mathrm{3}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{if}\:\:\mathrm{1}<{x}}\end{cases} \\ $$$${faind}\:\:\:\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}{f}\left({x}\right)=?\:\:{and}\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}{f}\left({x}\right)=? \\ $$ Answered by SEKRET last updated on 04/Jan/23 $$\:\boldsymbol{\mathrm{lim}}_{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{1}^{−} }…
Question Number 118681 by bemath last updated on 19/Oct/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{{x}}\:−\:\sqrt{\mathrm{sin}\:{x}}}{{x}^{\mathrm{2}} \:\sqrt{{x}}}\:=? \\ $$ Answered by benjo_mathlover last updated on 19/Oct/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}−\mathrm{sin}\:{x}}{{x}^{\mathrm{2}} \:\sqrt{{x}}}\:×\:\frac{\mathrm{1}}{\:\sqrt{{x}}\:+\sqrt{\mathrm{sin}\:{x}}}\:= \\…
Question Number 118427 by bramlexs22 last updated on 17/Oct/20 $$\:\:\:\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\sqrt[{{x}^{\mathrm{2}} \:}]{\mathrm{cos}\:{x}}\: \\ $$$$\:\:\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{1}+\mathrm{cos}\:\pi{x}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}} \\ $$ Answered by Dwaipayan Shikari last updated on…
Question Number 183936 by universe last updated on 31/Dec/22 Answered by Ar Brandon last updated on 31/Dec/22 $$\mathscr{L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\overset{{n}} {\:}{C}_{{k}} }\right)^{{n}} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{k}=\mathrm{1}}…
Question Number 183927 by greougoury555 last updated on 31/Dec/22 $$\:\underset{{x}\rightarrow\mathrm{2}\:} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\mathrm{2}\pi{x}\right)+\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}{x}\right)+\mathrm{tan}\:\left(\frac{\pi}{\mathrm{8}}{x}\right)}{{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{12}}\:=? \\ $$ Answered by cortano1 last updated on 31/Dec/22 $$\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\mathrm{2}\pi{x}\right)+\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}{x}\right)+\mathrm{tan}\:\left(\frac{\pi}{\mathrm{8}}{x}\right)}{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{6}\right)} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{2}}…
Question Number 118327 by bramlexs22 last updated on 17/Oct/20 $$\:\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{x}−\sqrt{\mathrm{2}−{x}^{\mathrm{2}} }}{\mathrm{2}{x}−\sqrt{\mathrm{2}+\mathrm{2}{x}^{\mathrm{2}} }}\:? \\ $$ Answered by TANMAY PANACEA last updated on 17/Oct/20 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}}…
Question Number 118272 by bramlexs22 last updated on 16/Oct/20 $$\:\:\:\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left(\frac{{a}}{\mathrm{1}−{x}^{{a}} }\:−\:\frac{{b}}{\mathrm{1}−{x}^{{b}} }\:\right)\:=\: \\ $$$${with}\:\left({a},{b}\right)\:\in\left(\mathbb{R}^{+} \right)^{\mathrm{2}} \: \\ $$ Answered by mathmax by abdo last…
Question Number 118247 by bemath last updated on 16/Oct/20 $$\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\mathrm{4}}]{\mathrm{1}+{x}^{\mathrm{4}} }\:−\:\sqrt[{\mathrm{5}}]{\mathrm{1}+{x}^{\mathrm{5}} }\:=? \\ $$ Answered by bobhans last updated on 16/Oct/20 $${solve}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\mathrm{4}}]{\mathrm{1}+{x}^{\mathrm{4}} }\:−\:\sqrt[{\mathrm{5}\:}]{\mathrm{1}+{x}^{\mathrm{5}}…
Question Number 118228 by bobhans last updated on 16/Oct/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sum_{\mathrm{r}=\mathrm{1}} ^{\mathrm{10}} \left(\mathrm{x}+\mathrm{r}\right)^{\mathrm{2020}} }{\left(\mathrm{x}^{\mathrm{1008}} +\mathrm{1}\right)\left(\mathrm{3x}^{\mathrm{1012}} +\mathrm{1}\right)}\:=?\: \\ $$ Answered by MJS_new last updated on 16/Oct/20…