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Category: Limits

Question-202161

Question Number 202161 by Calculusboy last updated on 22/Dec/23 Answered by som(math1967) last updated on 22/Dec/23 (m+1)!(1+3+5++2m+32m(m+2)(1+2+3++m+1)=(m+1)!m+22×2{1+(m+21)}2m(m+2)(m+1)(m+2)2=(m+1)!(m+2)2m(m+2)2(m+1)$$=\frac{{m}\left({m}+\mathrm{1}\right)×\left({m}−\mathrm{1}\right)!}{{m}\left({m}+\mathrm{1}\right)}…

Question-201680

Question Number 201680 by cortano12 last updated on 10/Dec/23 Answered by Calculusboy last updated on 11/Dec/23 Solution:substituteditectly,weget00(indeterminant)letp=2sinxsim2xdpdx=2cosx2cos2x$$\boldsymbol{{let}}\:\boldsymbol{{q}}=\boldsymbol{{sinx}}−\boldsymbol{{xcosx}}\:\:\:\frac{\boldsymbol{{dq}}}{\boldsymbol{{dx}}}=\boldsymbol{{cosx}}−\left(\boldsymbol{{cosx}}−\boldsymbol{{xsinx}}\right) \

Question-201221

Question Number 201221 by Calculusboy last updated on 02/Dec/23 Answered by MM42 last updated on 02/Dec/23 tan1atan1b=tan1(ab1+ab)$$\Rightarrow{tan}^{−\mathrm{1}} \left(\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}\right)−{tan}^{−\mathrm{1}} \left(\frac{{x}}{{x}+\mathrm{2}}\right)={tan}^{−\mathrm{1}} \left(\frac{{x}+\mathrm{2}}{\mathrm{2}{x}^{\mathrm{2}}…