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Author: Tinku Tara

Question-199467

Question Number 199467 by Calculusboy last updated on 04/Nov/23 Answered by cortano12 last updated on 04/Nov/23 $$\:\mathrm{4}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{a}−\mathrm{2}\right)\mathrm{x}^{\mathrm{3}} +\left(\mathrm{3}+\mathrm{c}\right)\mathrm{x}^{\mathrm{2}} +\left(\mathrm{b}−\mathrm{3}\right)\mathrm{x}+\mathrm{2}+\mathrm{d}}{\mathrm{x}^{\mathrm{2}} \:\left[\sqrt{\mathrm{1}+\frac{\mathrm{a}}{\mathrm{x}}+\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{b}}{\mathrm{x}^{\mathrm{3}} }+\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{4}} }}\:+\sqrt{\mathrm{1}+\frac{\mathrm{2}}{\mathrm{x}}−\frac{\mathrm{c}}{\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{3}}…

Question-199522

Question Number 199522 by cherokeesay last updated on 04/Nov/23 Answered by cortano12 last updated on 05/Nov/23 $$\left(\mathrm{g}_{\mathrm{1}} \right)\:\equiv\:\mathrm{y}=\mathrm{ax}+\mathrm{b}\:\left(\mathrm{tangent}\:\mathrm{line}\:\right) \\ $$$$\:\mathrm{where}\:\mathrm{a}\:=\:\frac{\mathrm{1}}{\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}\:=\:\mathrm{1} \\ $$$$\:\mathrm{and}\:\mathrm{b}\:=\:\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{8}}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\:\:\therefore\:\mathrm{y}=\mathrm{x}+\frac{\mathrm{1}}{\mathrm{8}} \\…

Find-1-lim-n-5n-25-3n-15-1-5n-2-lim-x-x-3-3x-2-1-1-3-x-3-3x-2-1-1-3-3-lim-x-0-cos-4x-3-1-sin-6-2x-

Question Number 199451 by hardmath last updated on 03/Nov/23 $$\mathrm{Find}: \\ $$$$\mathrm{1}.\:\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\mathrm{5}\boldsymbol{\mathrm{n}}}]{\frac{\mathrm{5n}\:−\:\mathrm{25}}{\mathrm{3n}\:+\:\mathrm{15}}} \\ $$$$\mathrm{2}.\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\infty} {\mathrm{lim}}\:\left(\sqrt[{\mathrm{3}}]{\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{3x}^{\mathrm{2}} \:+\:\mathrm{1}}\:−\:\sqrt[{\mathrm{3}}]{\mathrm{x}^{\mathrm{3}} \:−\:\mathrm{3x}^{\mathrm{2}} \:+\:\mathrm{1}}\:\right) \\ $$$$\mathrm{3}.\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\mathrm{4x}^{\mathrm{3}} \:−\:\mathrm{1}}{\mathrm{sin}^{\mathrm{6}} \:\mathrm{2x}}…

Question-199413

Question Number 199413 by universe last updated on 03/Nov/23 Answered by ajfour last updated on 03/Nov/23 $${b}\mathrm{cos}\:\alpha={R} \\ $$$$\left\{{R}\mathrm{tan}\:\alpha+\left(\frac{{a}}{{b}}\right){R}\right\}^{\mathrm{2}} +\left\{\left(\frac{{a}}{{b}}\right){R}\mathrm{tan}\:\alpha\right\}^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$${say}\:\:\:\mathrm{tan}\:\alpha={t} \\ $$$$\Rightarrow\:{t}^{\mathrm{2}}…