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Question-201680




Question Number 201680 by cortano12 last updated on 10/Dec/23
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$$\:\:\:\Subset \\ $$
Answered by Calculusboy last updated on 11/Dec/23
Solution: substitute ditectly, we get (0/0)(indeterminant)  let p=2sinx−sim2x     (dp/dx)=2cosx−2cos2x  let q=sinx−xcosx   (dq/dx)=cosx−(cosx−xsinx)  lim_(x→0)  (p/q)= lim_(x→0)  ((2cosx−2cos(2x))/(cosx−cosx+xsinx))=lim_(x→0) ((2cosx−2cos(2x))/(xsinx))  ((lim_(x→0) (2cosx−2cos(2x)))/(lim_(x→0) (xsinx)))=(0/0)  let u=2cosx−2cos(2x)       (du/dx)=−2sinx+4sin(2x)  let k=xsinx   (dk/dx)=sinx+xcosx  lim_(x→0)  (u/k)= lim_(x→0)  ((4sin(2x)−2sinx)/(sinx+xcosx))=(0/0)  let j=4sin(2x)−2sinx     (dj/dx)=8cos(2x)−2cosx  let g=sinx+xcosx     (dg/dx)=cosx+(cosx−xsinc)=2cosx−xsinx  lim_(x→0)  (j/g)= lim_(x→0)  ((8cos(2x)−2cosx)/(2cosx−xsinx))=((8−2)/(2−0))=(6/2)=3  ∴ lim_(x→0)  ((2sinx−sin(2x))/(sinx−xcosx))=3
$$\boldsymbol{{Solution}}:\:\boldsymbol{{substitute}}\:\boldsymbol{{ditectly}},\:\boldsymbol{{we}}\:\boldsymbol{{get}}\:\frac{\mathrm{0}}{\mathrm{0}}\left(\boldsymbol{{indeterminant}}\right) \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{p}}=\mathrm{2}\boldsymbol{{sinx}}−\boldsymbol{{sim}}\mathrm{2}\boldsymbol{{x}}\:\:\:\:\:\frac{\boldsymbol{{dp}}}{\boldsymbol{{dx}}}=\mathrm{2}\boldsymbol{{cosx}}−\mathrm{2}\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}} \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{q}}=\boldsymbol{{sinx}}−\boldsymbol{{xcosx}}\:\:\:\frac{\boldsymbol{{dq}}}{\boldsymbol{{dx}}}=\boldsymbol{{cosx}}−\left(\boldsymbol{{cosx}}−\boldsymbol{{xsinx}}\right) \\ $$$$\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\:\frac{\boldsymbol{{p}}}{\boldsymbol{{q}}}=\:\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\:\frac{\mathrm{2}\boldsymbol{{cosx}}−\mathrm{2}\boldsymbol{{cos}}\left(\mathrm{2}\boldsymbol{{x}}\right)}{\boldsymbol{{cosx}}−\boldsymbol{{cosx}}+\boldsymbol{{xsinx}}}=\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\mathrm{2}\boldsymbol{{cosx}}−\mathrm{2}\boldsymbol{{cos}}\left(\mathrm{2}\boldsymbol{{x}}\right)}{\boldsymbol{{xsinx}}} \\ $$$$\frac{\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\left(\mathrm{2}\boldsymbol{{cosx}}−\mathrm{2}\boldsymbol{{cos}}\left(\mathrm{2}\boldsymbol{{x}}\right)\right)}{\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\left(\boldsymbol{{xsinx}}\right)}=\frac{\mathrm{0}}{\mathrm{0}} \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{u}}=\mathrm{2}\boldsymbol{{cosx}}−\mathrm{2}\boldsymbol{{cos}}\left(\mathrm{2}\boldsymbol{{x}}\right)\:\:\:\:\:\:\:\frac{\boldsymbol{{du}}}{\boldsymbol{{dx}}}=−\mathrm{2}\boldsymbol{{sinx}}+\mathrm{4}\boldsymbol{{sin}}\left(\mathrm{2}\boldsymbol{{x}}\right) \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{k}}=\boldsymbol{{xsinx}}\:\:\:\frac{\boldsymbol{{dk}}}{\boldsymbol{{dx}}}=\boldsymbol{{sinx}}+\boldsymbol{{xcosx}} \\ $$$$\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\:\frac{\boldsymbol{{u}}}{\boldsymbol{{k}}}=\:\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\:\frac{\mathrm{4}\boldsymbol{{sin}}\left(\mathrm{2}\boldsymbol{{x}}\right)−\mathrm{2}\boldsymbol{{sinx}}}{\boldsymbol{{sinx}}+\boldsymbol{{xcosx}}}=\frac{\mathrm{0}}{\mathrm{0}} \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{j}}=\mathrm{4}\boldsymbol{{sin}}\left(\mathrm{2}\boldsymbol{{x}}\right)−\mathrm{2}\boldsymbol{{sinx}}\:\:\:\:\:\frac{\boldsymbol{{dj}}}{\boldsymbol{{dx}}}=\mathrm{8}\boldsymbol{{cos}}\left(\mathrm{2}\boldsymbol{{x}}\right)−\mathrm{2}\boldsymbol{{cosx}} \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{g}}=\boldsymbol{{sinx}}+\boldsymbol{{xcosx}}\:\:\:\:\:\frac{\boldsymbol{{dg}}}{\boldsymbol{{dx}}}=\boldsymbol{{cosx}}+\left(\boldsymbol{{cosx}}−\boldsymbol{{xsinc}}\right)=\mathrm{2}\boldsymbol{{cosx}}−\boldsymbol{{xsinx}} \\ $$$$\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\:\frac{\boldsymbol{{j}}}{\boldsymbol{{g}}}=\:\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\:\frac{\mathrm{8}\boldsymbol{{cos}}\left(\mathrm{2}\boldsymbol{{x}}\right)−\mathrm{2}\boldsymbol{{cosx}}}{\mathrm{2}\boldsymbol{{cosx}}−\boldsymbol{{xsinx}}}=\frac{\mathrm{8}−\mathrm{2}}{\mathrm{2}−\mathrm{0}}=\frac{\mathrm{6}}{\mathrm{2}}=\mathrm{3} \\ $$$$\therefore\:\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\:\frac{\mathrm{2}\boldsymbol{{sinx}}−\boldsymbol{{sin}}\left(\mathrm{2}\boldsymbol{{x}}\right)}{\boldsymbol{{sinx}}−\boldsymbol{{xcosx}}}=\mathrm{3}\: \\ $$$$ \\ $$

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