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Question Number 156421 by cortano last updated on 11/Oct/21 Commented by MJS_new last updated on 11/Oct/21 $$\mathrm{2} \\ $$ Commented by cortano last updated on…
Question Number 25300 by SAGARSTARK last updated on 07/Dec/17 Commented by prakash jain last updated on 08/Dec/17 $$\mathrm{You}\:\mathrm{may}\:\mathrm{also}\:\mathrm{want}\:\mathrm{to}\:\mathrm{see} \\ $$$$\mathrm{Q2088}.\:\mathrm{Different}\:\mathrm{but}\:\mathrm{similar} \\ $$$$\mathrm{question}. \\ $$ Commented…
Question Number 25287 by 9931558112 last updated on 07/Dec/17 $${limit}_{{x}−\mathrm{0}^{−} } \mathrm{sin}\:\left({log}_{{e}} {x}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 90788 by jagoll last updated on 26/Apr/20 $$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\left(\frac{{x}+\mathrm{2}}{{x}+\mathrm{1}}\right)^{\frac{{x}}{\mathrm{2}}} \\ $$ Commented by mathmax by abdo last updated on 26/Apr/20 $${let}\:{f}\left({x}\right)=\left(\frac{{x}+\mathrm{2}}{{x}+\mathrm{1}}\right)^{\frac{{x}}{\mathrm{2}}} \:\:\Rightarrow{f}\left({x}\right)\:=_{{x}=−{t}} \:\:\left(\frac{−{t}+\mathrm{2}}{−{t}+\mathrm{1}}\right)^{−\frac{{t}}{\mathrm{2}}}…
Question Number 25221 by chernoaguero@gmail.com last updated on 06/Dec/17 $$\mathrm{Find}\:\mathrm{the}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\:\frac{\mathrm{e}^{\mathrm{x}} }{\:\sqrt{\mathrm{e}^{\mathrm{2x}} +\mathrm{1}}} \\ $$ Answered by jota+ last updated on 06/Dec/17 $$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{1}/{e}^{{x}} \right){e}^{{x}}…
Question Number 90722 by jagoll last updated on 25/Apr/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\sqrt{\mathrm{1}+{x}}\:−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} }{\mathrm{cos}\:{x}−\mathrm{1}}\:=\:? \\ $$ Commented by john santu last updated on 25/Apr/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\sqrt{\mathrm{1}+{x}}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}}…
Question Number 156250 by cortano last updated on 09/Oct/21 $$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{x}\:\sqrt{\mathrm{cos}\:\mathrm{2x}}\:\sqrt[{\mathrm{3}}]{\mathrm{cos}\:\mathrm{3x}}\:\sqrt[{\mathrm{4}}]{\mathrm{cos}\:\mathrm{4x}}}{\mathrm{x}^{\mathrm{2}} }\:=? \\ $$ Commented by john_santu last updated on 09/Oct/21 $${limit}=\frac{\mathrm{4}×\mathrm{5}}{\mathrm{4}}=\mathrm{5} \\ $$ Terms…
Question Number 25068 by NECx last updated on 02/Dec/17 $${Evaluate}\: \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\left(\frac{{tan}\mathrm{2}{x}}{{x}−\pi/\mathrm{2}}\right) \\ $$ Answered by jota+ last updated on 03/Dec/17 $${Use}\:{Hopital}\:{rule} \\ $$$$=\frac{\underset{{x}\rightarrow\pi/\mathrm{2}}…
Question Number 156108 by cortano last updated on 08/Oct/21 $$\:\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{8}}} {\mathrm{lim}}\:\frac{\mathrm{1}+\mathrm{cot}\:\mathrm{6x}}{\mathrm{1}−\mathrm{sin}\:\mathrm{4x}}\:=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com