Question Number 141901 by gsk2684 last updated on 24/May/21 Commented by gsk2684 last updated on 24/May/21 $${does}\:{this}\:{series}\:{converges}\:{or}\:{diverges} \\ $$$${by}\:{using}\:{comparison}\:{test}\:{or}\:{limit} \\ $$$${comparison}\:{test}?\:{explanation}\:{please} \\ $$ Terms of…
Question Number 141900 by iloveisrael last updated on 24/May/21 $$\:\mathcal{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{x}} \:\mathrm{cos}\:{x}−{x}−\mathrm{1}}{{x}^{\mathrm{3}} }\:=? \\ $$ Answered by Dwaipayan Shikari last updated on 24/May/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}}…
Question Number 10830 by Saham last updated on 26/Feb/17 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{3}^{\mathrm{x}} \:−\:\mathrm{3}^{−\mathrm{x}} }{\mathrm{3}^{\mathrm{x}} \:+\:\mathrm{3}^{−\mathrm{x}} } \\ $$ Answered by bahmanfeshki last updated on 26/Feb/17 $$\underset{{x}\rightarrow\infty}…
Question Number 141885 by mnjuly1970 last updated on 24/May/21 $$\:\:\: \\ $$$$\:\:\:\:\:\:{easy}\:\:{question}: \\ $$$$\:\:\:\:{if}\:\:\:{lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{1}−{cos}\left(\mathrm{1}−{cos}\left(\mathrm{1}−{cos}\left({x}\right)\right)\right)}{{x}^{\mathrm{8}} }\:=\mathrm{2}^{\:{a}} \\ $$$$\:\:\:\:\:\:\:\:{then}\:\:\:{a}=??\:\: \\ $$ Answered by Dwaipayan Shikari last…
Question Number 141858 by gsk2684 last updated on 24/May/21 Commented by gsk2684 last updated on 24/May/21 $${solution}\:{please} \\ $$ Answered by Dwaipayan Shikari last updated…
Question Number 141779 by iloveisrael last updated on 23/May/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{x}^{\mathrm{2}} }{{x}+\mathrm{2}−\mathrm{2}\sqrt{\mathrm{1}+{x}}}\:=? \\ $$ Answered by Willson last updated on 23/May/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{x}+\mathrm{2}−\mathrm{2}\sqrt{\mathrm{1}+\mathrm{x}}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2x}}{\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{x}}}}…
Question Number 76191 by abdomathmax last updated on 25/Dec/19 $${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\frac{{e}^{{x}} −{e}^{\left[{x}\right]} }{{x}} \\ $$ Answered by Rio Michael last updated on 25/Dec/19 $${Am}\:{not}\:{sure}\:{if}\:{this}\:{limit} \\…
Question Number 141699 by cesarL last updated on 22/May/21 $${If}\:{lim}_{{x}\rightarrow\mathrm{1}} =\frac{\sqrt[{{k}}]{{x}}−\mathrm{1}}{{x}−\mathrm{1}}={L}\neq\mathrm{0}\:\:{Find}\:{lim}_{{x}\rightarrow\mathrm{0}} \frac{\sqrt{{x}+\mathrm{1}}−\mathrm{1}}{\:\sqrt[{{k}}]{{x}+\mathrm{1}}−\mathrm{1}} \\ $$ Answered by iloveisrael last updated on 22/May/21 $${Given}\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt[{{k}\:}]{{x}}−\mathrm{1}}{{x}−\mathrm{1}}\:=\:{L}\:{equivalent} \\ $$$${to}\:\underset{{t}\rightarrow\mathrm{1}}…
Question Number 76151 by Rio Michael last updated on 24/Dec/19 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{{k}} {e}^{−\mathrm{4}{x}} ,\:{k}>\mathrm{0} \\ $$ Commented by kaivan.ahmadi last updated on 24/Dec/19 $${lim}_{{x}−\rightarrow\infty} \frac{{x}^{{k}}…
Question Number 141627 by cherokeesay last updated on 21/May/21 Answered by MJS_new last updated on 21/May/21 $$\frac{\mathrm{sin}\:\mathrm{2}{x}\:+\mathrm{sin}\:\mathrm{4}{x}}{\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}\:+\mathrm{cos}\:\mathrm{4}{x}}=\frac{−\mathrm{2sin}\:{x}\:\mathrm{cos}\:{x}\:\left(\mathrm{1}−\mathrm{4cos}^{\mathrm{2}} \:{x}\right)}{\mathrm{1}+\mathrm{2cos}^{\mathrm{2}} \:{x}\:\left(\mathrm{1}−\mathrm{4sin}^{\mathrm{2}} \:{x}\right)}= \\ $$$$\:\:\:\:\:\left[\mathrm{sin}\:{x}\:=\frac{\mathrm{tan}\:{x}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}}}\wedge\mathrm{cos}\:{x}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}}}\right] \\…