Question-203694

Question Number 203694 by Numsey last updated on 26/Jan/24

Answered by Calculusboy last updated on 26/Jan/24

Solution: by sub directly,we get (0/0)(indeterminant) let 𝚫=lim_(x→0) (((1+x)^5 )/x^2 )−lim_(x→0) (e^(5x) /x^2 )+lim_(x→0) ((sin(x^2 ))/x^2 ) 𝚫=lim_(x→0) ((5(1+x)^4 )/(2x))−lim_(x→0) ((5e^(5x) )/(2x))+1 𝚫=lim_(x→0) ((20(1+x)^3 )/2)−lim_(x→0) ((25e^(5x) )/2)+1 𝚫=10lim_(x→0) (1+x)^3 −((25)/2)lim_(x→0) e^(5x) +1 𝚫=10−((25)/2)+1 𝚫=−(3/2)=−1.5 ∴lim_(x→0) (((1+x)^5 −e^(5x) +sinx^2 )/x^2 )=−(3/2)=−1.5

$$\boldsymbol{{Solution}}:\:\boldsymbol{{by}}\:\boldsymbol{{sub}}\:\boldsymbol{{directly}},\boldsymbol{{we}}\:\boldsymbol{{get}}\:\frac{\mathrm{0}}{\mathrm{0}}\left(\boldsymbol{{indeterminant}}\right) \\ $$$$\boldsymbol{{let}}\:\boldsymbol{\Delta}=\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\left(\mathrm{1}+\boldsymbol{{x}}\right)^{\mathrm{5}} }{\boldsymbol{{x}}^{\mathrm{2}} }−\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\boldsymbol{{e}}^{\mathrm{5}\boldsymbol{{x}}} }{\boldsymbol{{x}}^{\mathrm{2}} }+\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\boldsymbol{{sin}}\left(\boldsymbol{{x}}^{\mathrm{2}} \right)}{\boldsymbol{{x}}^{\mathrm{2}} } \\ $$$$\boldsymbol{\Delta}=\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\mathrm{5}\left(\mathrm{1}+\boldsymbol{{x}}\right)^{\mathrm{4}} }{\mathrm{2}\boldsymbol{{x}}}−\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\mathrm{5}\boldsymbol{{e}}^{\mathrm{5}\boldsymbol{{x}}} }{\mathrm{2}\boldsymbol{{x}}}+\mathrm{1} \\ $$$$\boldsymbol{\Delta}=\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\mathrm{20}\left(\mathrm{1}+\boldsymbol{{x}}\right)^{\mathrm{3}} }{\mathrm{2}}−\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\mathrm{25}\boldsymbol{{e}}^{\mathrm{5}\boldsymbol{{x}}} }{\mathrm{2}}+\mathrm{1} \\ $$$$\boldsymbol{\Delta}=\mathrm{10}\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\left(\mathrm{1}+\boldsymbol{{x}}\right)^{\mathrm{3}} −\frac{\mathrm{25}}{\mathrm{2}}\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}e}}^{\mathrm{5}\boldsymbol{{x}}} +\mathrm{1} \\ $$$$\boldsymbol{\Delta}=\mathrm{10}−\frac{\mathrm{25}}{\mathrm{2}}+\mathrm{1} \\ $$$$\boldsymbol{\Delta}=−\frac{\mathrm{3}}{\mathrm{2}}=−\mathrm{1}.\mathrm{5} \\ $$$$\therefore\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\left(\mathrm{1}+\boldsymbol{{x}}\right)^{\mathrm{5}} −\boldsymbol{{e}}^{\mathrm{5}\boldsymbol{{x}}} +\boldsymbol{{sinx}}^{\mathrm{2}} }{\boldsymbol{{x}}^{\mathrm{2}} }=−\frac{\mathrm{3}}{\mathrm{2}}=−\mathrm{1}.\mathrm{5} \\ $$

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