Question Number 203374 by otchereabdullai@gmail.com last updated on 17/Jan/24
Answered by Calculusboy last updated on 18/Jan/24
$$\boldsymbol{{Solution}}:\:\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\boldsymbol{{tan}}\mathrm{5}\boldsymbol{{x}}}{\mathrm{3}\boldsymbol{{x}}}\:\:\:\left(\boldsymbol{{by}}\:\boldsymbol{{using}}\:\boldsymbol{{algebraic}}\:\boldsymbol{{methods}}\right) \\ $$$$\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\boldsymbol{{tan}}\mathrm{5}\boldsymbol{{x}}}{\mathrm{3}\boldsymbol{{x}}}×\frac{\mathrm{5}\boldsymbol{{x}}}{\mathrm{5}\boldsymbol{{x}}}=\frac{\mathrm{5}\boldsymbol{{x}}}{\mathrm{3}\boldsymbol{{x}}}×\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\boldsymbol{{tan}}\mathrm{5}\boldsymbol{{x}}}{\mathrm{5}\boldsymbol{{x}}} \\ $$$$\boldsymbol{{NB}}:\:\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\boldsymbol{{tanax}}}{\boldsymbol{{x}}}=\mathrm{1}\:\:\boldsymbol{{then}}\:\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\boldsymbol{{tan}}\mathrm{5}\boldsymbol{{x}}}{\mathrm{5}\boldsymbol{{x}}}=\mathrm{1} \\ $$$$\therefore\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\boldsymbol{{tan}}\mathrm{5}\boldsymbol{{x}}}{\mathrm{3}\boldsymbol{{x}}}=\frac{\mathrm{5}}{\mathrm{3}}×\mathrm{1}=\frac{\mathrm{5}}{\mathrm{3}} \\ $$
Answered by Calculusboy last updated on 18/Jan/24
![Solution: by using BF methods let u=x u′=1 u′′=0 & v=sinx v_1 =−cosx v_2 =−sinx v_3 =cosx I=uv_1 −u′v_2 +u′′v_3 −∙∙∙ I=−∣xcosx∣_0 ^𝛑 +∣sinx∣_0 ^𝛑 +C I=−[𝛑cos(𝛑)−0∙cos(0)∣+[sin(𝛑)−sin(0)] I=− (−𝛑−0)+(0−0) I=𝛑](https://www.tinkutara.com/question/Q203381.png)
$$\boldsymbol{{Solution}}:\:\boldsymbol{{by}}\:\boldsymbol{{using}}\:\boldsymbol{{BF}}\:\boldsymbol{{methods}} \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{u}}=\boldsymbol{{x}}\:\boldsymbol{{u}}'=\mathrm{1}\:\boldsymbol{{u}}''=\mathrm{0}\:\:\&\:\boldsymbol{{v}}=\boldsymbol{{sinx}}\:\boldsymbol{{v}}_{\mathrm{1}} =−\boldsymbol{{cosx}}\:\boldsymbol{{v}}_{\mathrm{2}} =−\boldsymbol{{sinx}}\:\boldsymbol{{v}}_{\mathrm{3}} =\boldsymbol{{cosx}} \\ $$$$\boldsymbol{{I}}=\boldsymbol{{uv}}_{\mathrm{1}} −\boldsymbol{{u}}'\boldsymbol{{v}}_{\mathrm{2}} +\boldsymbol{{u}}''\boldsymbol{{v}}_{\mathrm{3}} −\centerdot\centerdot\centerdot \\ $$$$\boldsymbol{{I}}=−\mid\boldsymbol{{xcosx}}\mid_{\mathrm{0}} ^{\boldsymbol{\pi}} +\mid\boldsymbol{{sinx}}\mid_{\mathrm{0}} ^{\boldsymbol{\pi}} +\boldsymbol{{C}} \\ $$$$\boldsymbol{{I}}=−\left[\boldsymbol{\pi{cos}}\left(\boldsymbol{\pi}\right)−\mathrm{0}\centerdot\boldsymbol{{cos}}\left(\mathrm{0}\right)\mid+\left[\boldsymbol{{sin}}\left(\boldsymbol{\pi}\right)−\boldsymbol{{sin}}\left(\mathrm{0}\right)\right]\right. \\ $$$$\boldsymbol{{I}}=−\:\left(−\boldsymbol{\pi}−\mathrm{0}\right)+\left(\mathrm{0}−\mathrm{0}\right) \\ $$$$\boldsymbol{{I}}=\boldsymbol{\pi} \\ $$
Answered by Calculusboy last updated on 18/Jan/24
$$\boldsymbol{{Solution}}:\:\left(\mathrm{5}\boldsymbol{{B}}\right)\: \\ $$$$\left.\boldsymbol{{a}}\right)\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{1}} {\boldsymbol{{m}}}\frac{\mathrm{1}−\boldsymbol{{x}}}{\mathrm{1}+\boldsymbol{{x}}}=\frac{\mathrm{1}−\mathrm{1}}{\mathrm{1}+\mathrm{1}}=\frac{\mathrm{0}}{\mathrm{2}}=\mathrm{0}\:\: \\ $$$$\left(\boldsymbol{{by}}\:\boldsymbol{{sub}}\:\boldsymbol{{directly}},\boldsymbol{{L}}'\boldsymbol{{hospital}}\:\boldsymbol{{rule}}\:\boldsymbol{{is}}\:\boldsymbol{{discarded}}\right) \\ $$$$\left.\boldsymbol{{b}}\right)\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{1}} {\boldsymbol{{m}}}\frac{\mathrm{1}+\boldsymbol{{cosx}\pi}}{\mathrm{1}−\boldsymbol{{x}}}=\frac{\mathrm{1}+\boldsymbol{{cos}\pi}}{\mathrm{1}−\mathrm{1}}=\frac{\mathrm{1}−\mathrm{1}}{\mathrm{1}−\mathrm{1}}=\frac{\mathrm{0}}{\mathrm{0}} \\ $$$$\left(\boldsymbol{{indeterminant}},\boldsymbol{{by}}\:\boldsymbol{{using}}\:\boldsymbol{{L}}'\boldsymbol{{hospital}}\:\boldsymbol{{rule}}\right) \\ $$
Commented by otchereabdullai@gmail.com last updated on 18/Jan/24
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