Question Number 155829 by zainaltanjung last updated on 05/Oct/21 $$\mathrm{Find}\:\mathrm{this}\:\mathrm{excercise}\:\mathrm{about}\:\mathrm{limits} \\ $$$$\mathrm{trigonometri} \\ $$$$\left.\mathrm{1}\right).\:\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2cos}\:\theta−\mathrm{2}}{\mathrm{3}\theta} \\ $$$$\left.\mathrm{2}\right).\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{x}}{\mathrm{x}^{\frac{\mathrm{2}}{\mathrm{3}}} } \\ $$$$\left.\mathrm{3}\right).\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{4t}^{\mathrm{2}} +\mathrm{3t}\:\mathrm{sin}\:\mathrm{t}}{\mathrm{t}^{\mathrm{2}} } \\…
Question Number 24755 by Anoop kumar last updated on 25/Nov/17 $${Given} \\ $$$${f}\left({x}\right)\:=\underset{{x}=\mathrm{1}} {\overset{{n}} {\sum}}{tan}\left(\frac{{x}}{\mathrm{2}^{{r}} }\right).{sec}\left(\frac{{x}}{\mathrm{2}^{{r}−\mathrm{1}} }\right)\: \\ $$$$\:\:\:\:\:\:\:\:\:{where}\:{r}\:{and}\:{n}\:\varepsilon{N} \\ $$$${g}\left({x}\right)\:=\underset{{n}\rightarrow\propto} {\mathrm{li}{m}}\:\:\frac{{ln}\left({f}\left({x}\right)+{tan}\frac{{x}}{\mathrm{2}^{{n}} }\right)\:−\left({f}\left({x}\right)+{tan}\frac{{x}}{\mathrm{2}^{{n}} }\right).\left[{sin}\left({tan}\frac{{x}}{\mathrm{2}}\right)\right.}{\mathrm{1}+\left({f}\left({x}\right)\:\:+\:\:{tan}\frac{{x}}{\mathrm{2}^{{n}} }\right)^{{n}}…
Question Number 90281 by jagoll last updated on 22/Apr/20 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{1}−\sqrt{\mathrm{x}}}{\left(\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{x}\right)\right)^{\mathrm{2}} }\:=\:? \\ $$ Commented by jagoll last updated on 22/Apr/20 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{x}}}}{\mathrm{2cos}^{−\mathrm{1}} \left(\mathrm{x}\right).\left(\frac{−\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}}…
Question Number 155812 by zainaltanjung last updated on 05/Oct/21 $$\mathrm{Use}\:\mathrm{algebraic}\:\mathrm{simplifications}\:\mathrm{to}\:\mathrm{help} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{limits},\:\mathrm{if}\:\mathrm{they}\:\mathrm{exist}. \\ $$$$\left.\mathrm{1}\right).\:\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{4}}{\mathrm{x}−\mathrm{2}} \\ $$$$\left.\mathrm{2}\right).\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\:\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{x}}{\mathrm{2x}^{\mathrm{2}} +\mathrm{5x}−\mathrm{7}} \\ $$$$\left.\mathrm{3}\right).\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\:\:\frac{\mathrm{3x}^{\mathrm{2}} −\mathrm{13x}−\mathrm{10}}{\mathrm{2x}^{\mathrm{2}}…
Question Number 90226 by manuel__ last updated on 22/Apr/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{x}^{\mathrm{3}} \left({e}^{{x}} −\mathrm{1}\right)−{x}^{\mathrm{2}} }{{xe}^{{x}} −{x}}\right)=? \\ $$ Commented by john santu last updated on 22/Apr/20…
Question Number 90210 by manuel__ last updated on 22/Apr/20 $$\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {\mathrm{lim}}\:\left(\frac{\mathrm{1}−\mathrm{2cos}\:\left({x}\right)}{\pi−\mathrm{3}{x}}\right)=? \\ $$ Commented by mathmax by abdo last updated on 22/Apr/20 $${f}\left({x}\right)=\frac{\mathrm{2}{cosx}−\mathrm{1}}{\mathrm{3}{x}−\pi}\:=\frac{\mathrm{2}{cosx}−\mathrm{1}}{\mathrm{3}\left({x}−\frac{\pi}{\mathrm{3}}\right)}\:{we}\:{do}\:{the}\:{changement}\:{x}−\frac{\pi}{\mathrm{3}}={t}\:\Rightarrow \\ $$$$\left({x}\rightarrow\frac{\pi}{\mathrm{3}}\Leftrightarrow\:{t}\rightarrow\mathrm{0}\right)\:{and}\:{f}\left({x}\right)=\frac{\mathrm{2}{cos}\left({t}+\frac{\pi}{\mathrm{3}}\right)−\mathrm{1}}{\mathrm{3}{t}}={g}\left({t}\right)…
Question Number 90151 by JosephK last updated on 21/Apr/20 $${find}\:{the}\:{limit}\:{of}\: \\ $$$${lim}\:\:\frac{\mathrm{1}}{{t}\left(\sqrt{\mathrm{1}+{t}}\right.}−\frac{\mathrm{1}}{{t}} \\ $$$${t}\rightarrow\mathrm{0} \\ $$ Commented by abdomathmax last updated on 21/Apr/20 $${f}\left({t}\right)=\frac{\mathrm{1}}{{t}\sqrt{\mathrm{1}+{t}}}−\frac{\mathrm{1}}{{t}}\:\Rightarrow{f}\left({t}\right)=\frac{{t}−{t}\sqrt{\mathrm{1}+{t}}}{{t}^{\mathrm{2}} \sqrt{\mathrm{1}+{t}}}…
Question Number 155642 by mathlove last updated on 03/Oct/21 Commented by cortano last updated on 03/Oct/21 $$\mathrm{8}\sqrt{\mathrm{2}}\:\mathrm{ln}\:^{\mathrm{2}} \left(\mathrm{3}\right) \\ $$ Commented by mathlove last updated…
Question Number 155639 by cortano last updated on 03/Oct/21 Commented by yeti123 last updated on 03/Oct/21 $$\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{1}−\mathrm{cos}\:\theta}\:=\:\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2sin}^{\mathrm{2}} \left(\theta/\mathrm{2}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\underset{\theta\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2}\theta}×\frac{\left(\theta/\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{2sin}^{\mathrm{2}} \left(\theta/\mathrm{2}\right)}×\frac{\mathrm{2}\theta}{\left(\theta/\mathrm{2}\right)^{\mathrm{2}}…
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