Question Number 110988 by Rio Michael last updated on 01/Sep/20 $$\:\mathrm{Evaluate}\:\mathrm{without}\:\mathrm{using}\:\mathrm{L}'\mathrm{hopital}'\mathrm{s}\:\mathrm{rule} \\ $$$$\:\:\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\:\frac{\sqrt{{x}}−\mathrm{2}}{{x}−\mathrm{4}} \\ $$ Answered by Rasheed.Sindhi last updated on 01/Sep/20 $$\:\:\underset{{x}\rightarrow\mathrm{4}} {\mathrm{lim}}\:\frac{\sqrt{{x}}−\mathrm{2}}{{x}−\mathrm{4}}…
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Question Number 176486 by cortano1 last updated on 20/Sep/22 $$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\left[\:\frac{\mathrm{tan}\:\mathrm{x}−\mathrm{x}}{\mathrm{x}^{\mathrm{5}} }\:−\frac{\mathrm{1}}{\mathrm{3x}^{\mathrm{2}} }\:\right]=? \\ $$ Answered by blackmamba last updated on 20/Sep/22 $$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{3}\left(\mathrm{tan}\:{x}−{x}\right)−{x}^{\mathrm{3}} }{\mathrm{3}{x}^{\mathrm{5}}…
Question Number 110869 by Study last updated on 31/Aug/20 Answered by mathdave last updated on 31/Aug/20 $${let}\:{I}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{x}!^{\frac{\mathrm{1}}{{x}}} \\ $$$${by}\:{loggin}\:{both}\:{side}\:{we}\:{have} \\ $$$$\mathrm{ln}{I}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}{x}!}{{x}}\:\:\:\:\:\:{but}\:\:{x}!=\Gamma\left({x}+\mathrm{1}\right) \\ $$$$\mathrm{ln}{I}=\underset{{x}\rightarrow\mathrm{0}}…
Question Number 110861 by apriani last updated on 31/Aug/20 Answered by bemath last updated on 31/Aug/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt{\mathrm{x}^{\mathrm{3}} −\mathrm{6x}^{\mathrm{2}} }−\left(\mathrm{x}+\mathrm{2}\right)= \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{x}^{\mathrm{3}} −\mathrm{6x}^{\mathrm{2}} \right)−\left(\mathrm{x}^{\mathrm{2}}…
Question Number 176394 by Matica last updated on 18/Sep/22 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}−{cos}\sqrt{\mid{x}\mid}\right)^{\mathrm{2}} }{\mathrm{1}−\sqrt{{cosx}}}\:=\:? \\ $$ Commented by cortano1 last updated on 18/Sep/22 $$\:\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\:\frac{\left(\mathrm{1}−\mathrm{cos}\:\sqrt{\mid\mathrm{x}\mid}\:\right)^{\mathrm{2}} }{\mathrm{1}−\sqrt{\mathrm{cos}\:\mathrm{x}}}\:\left(\mathrm{DNE}\right)…
Question Number 176375 by cortano1 last updated on 17/Sep/22 $$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\mathrm{2cos}\:^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{x}}\right)−\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{x}}\right)+\mathrm{3}}{\mathrm{x}+\sqrt{\mathrm{x}}}\:=?\: \\ $$ Commented by a.lgnaoui last updated on 18/Sep/22 $${posons}\:\:{X}=\frac{\mathrm{1}}{{x}}\:\:\:{x}=\frac{\mathrm{1}}{{X}} \\ $$$$\left[\frac{\mathrm{5}−\mathrm{sin}\:^{\mathrm{2}}…
Question Number 110809 by Study last updated on 30/Aug/20 $${li}\underset{{x}\rightarrow\infty} {{m}}\sqrt{{x}!}=? \\ $$ Commented by Her_Majesty last updated on 30/Aug/20 $$\infty \\ $$ Commented by…
Question Number 110760 by bobhans last updated on 30/Aug/20 Answered by john santu last updated on 30/Aug/20 $$\begin{cases}{{x}+\mathrm{1}={m}}\\{{m}\rightarrow\mathrm{0}}\end{cases} \\ $$$${L}=\underset{{m}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{2}{m}−\mathrm{tan}\:{m}}{{m}^{\mathrm{3}} } \\ $$$${L}=\:\underset{{m}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:{m}\:\mathrm{cos}\:{m}−\mathrm{tan}\:{m}}{{m}^{\mathrm{3}}…
Question Number 110736 by abdullahquwatan last updated on 30/Aug/20 $$\mathrm{given}\:{f}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+{c}\:\mathrm{and}\:{f}\left({x}\right) \\ $$$$\mathrm{is}\:\mathrm{negative}\:\mathrm{on}\:{a}<{x}<{b}.\:\mathrm{The}\:\mathrm{value}\:\mathrm{of} \\ $$$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\frac{\left({x}−{a}\right)^{\mathrm{2}} }{\mathrm{1}−\mathrm{cos}\:\left({f}\left({x}\right)\right)} \\ $$ Commented by abdullahquwatan last updated on…