Question Number 128607 by john_santu last updated on 08/Jan/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{x}.\mathrm{cot}\:\mathrm{x}−\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=? \\ $$ Answered by malwan last updated on 08/Jan/21 $$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{\frac{{x}}{{tan}\:{x}}\:−\:\mathrm{1}}{{x}^{\mathrm{2}} }\:=\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{{x}−{tan}\:{x}}{{x}^{\mathrm{2}}…
Question Number 128605 by john_santu last updated on 08/Jan/21 $$\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\sqrt{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} }−\sqrt{\mathrm{1}−\mathrm{x}+\mathrm{x}^{\mathrm{2}} }=? \\ $$ Commented by john_santu last updated on 08/Jan/21 Answered by mathmax…
Question Number 128604 by john_santu last updated on 08/Jan/21 $$\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{x}} −\mathrm{x}}{\mathrm{ln}\:\mathrm{x}−\mathrm{x}+\mathrm{1}}\:? \\ $$ Answered by liberty last updated on 09/Jan/21 $$\:\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{x}} \right)=\mathrm{x}^{\mathrm{x}} \left(\mathrm{ln}\:\mathrm{x}+\mathrm{1}\right) \\…
Question Number 128593 by Eric002 last updated on 08/Jan/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mid\mathrm{3}{x}−\mathrm{1}\mid−\mid\mathrm{3}{x}+\mathrm{1}\mid}{{x}} \\ $$ Commented by MJS_new last updated on 08/Jan/21 $$\frac{\mid\mathrm{3}{x}−\mathrm{1}\mid−\mid\mathrm{3}{x}+\mathrm{1}\mid}{{x}}=\frac{\mathrm{3}}{{x}}\left(\mid{x}−\frac{\mathrm{1}}{\mathrm{3}}\mid−\mid{x}+\frac{\mathrm{1}}{\mathrm{3}}\mid\right)= \\ $$$$\:\:\:\:\:\left[−\frac{\mathrm{1}}{\mathrm{3}}\leqslant{x}\leqslant\frac{\mathrm{1}}{\mathrm{3}}\right] \\ $$$$=\frac{\mathrm{3}}{{x}}\left(−\mathrm{2}{x}\right)=−\frac{\mathrm{6}{x}}{{x}}=−\mathrm{6}\forall{x}\in\left[−\frac{\mathrm{1}}{\mathrm{3}};\:\frac{\mathrm{1}}{\mathrm{3}}\right]\wedge{x}\neq\mathrm{0}…
Question Number 63054 by jimful last updated on 28/Jun/19 $${if}\:\Sigma\mid{a}_{{n}} \:\mid\:{is}\:{convergent},\:{then} \\ $$$${prove}\:{that}\:{there}\:{exists}\: \\ $$$${a}\:{subsequence}\:\left\{{n}_{{k}} {a}_{{n}_{{k}} } \right\}\:\:{with} \\ $$$$\underset{{k}\rightarrow\infty} {\mathrm{lim}}{n}_{{k}} {a}_{{n}_{{k}} } =\mathrm{0} \\…
Question Number 128567 by mnjuly1970 last updated on 08/Jan/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:…{nice}\:\:\:{calculus}… \\ $$$$\:{prove}\:{that}:::: \\ $$$$\:\:\:\phi=\int_{\mathrm{0}} ^{\:\infty} {sin}\left({x}^{\mathrm{2}} \right){sin}\left({x}^{−\mathrm{2}} \right){dx} \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\frac{\pi}{\mathrm{2}}}\:\left({e}^{−\mathrm{2}} +{sin}\left(\mathrm{2}\right)−{cos}\left(\mathrm{2}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$…
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Question Number 128456 by Study last updated on 07/Jan/21 $${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\left({cosx}\right)^{{logx}} =??? \\ $$ Commented by mr W last updated on 07/Jan/21 $$\rightarrow\mathrm{1}^{−\infty} =\mathrm{1} \\…
Question Number 128439 by Study last updated on 07/Jan/21 $${li}\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {{m}}\frac{{cosx}}{\mathrm{1}−{sinx}}=??? \\ $$ Answered by benjo_mathlover last updated on 07/Jan/21 $$\mathrm{let}\:\mathrm{x}=\frac{\pi}{\mathrm{2}}+\mathrm{z}\:\wedge\:\mathrm{z}\rightarrow\mathrm{0} \\ $$$$\:\underset{\mathrm{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}+\mathrm{z}\right)}{\mathrm{1}−\mathrm{sin}\:\left(\mathrm{z}+\frac{\pi}{\mathrm{2}}\right)}=\underset{\mathrm{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{sin}\:\mathrm{z}}{\mathrm{1}−\mathrm{cos}\:\mathrm{z}}…
Question Number 128411 by bramlexs22 last updated on 07/Jan/21 $$\:\mathrm{If}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{ln}\:\left({a}+{x}\right)−\mathrm{ln}\:{a}}{{x}}\right)+\:{k}\:\underset{{x}\rightarrow\mathrm{e}} {\mathrm{lim}}\left(\frac{\mathrm{ln}\:{x}−\mathrm{1}}{{x}−{e}}\right)=\mathrm{1} \\ $$$$\mathrm{then}\:{k}\:=\: \\ $$ Answered by Dwaipayan Shikari last updated on 07/Jan/21 $$\underset{{x}\rightarrow\mathrm{0}}…