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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_(xβ†’0) (((1+x)^5 )/x^2 )βˆ’lim_(xβ†’0) (e^(5x) /x^2 )+lim_(xβ†’0) ((sin(x^2 ))/x^2 )  𝚫=lim_(xβ†’0) ((5(1+x)^4 )/(2x))βˆ’lim_(xβ†’0) ((5e^(5x) )/(2x))+1  𝚫=lim_(xβ†’0) ((20(1+x)^3 )/2)βˆ’lim_(xβ†’0) ((25e^(5x) )/2)+1  𝚫=10lim_(xβ†’0) (1+x)^3 βˆ’((25)/2)lim_(xβ†’0) e^(5x) +1  𝚫=10βˆ’((25)/2)+1  𝚫=βˆ’(3/2)=βˆ’1.5  ∴lim_(xβ†’0) (((1+x)^5 βˆ’e^(5x) +sinx^2 )/x^2 )=βˆ’(3/2)=βˆ’1.5
Solution:bysubdirectly,weget00(indeterminant)letΞ”=limxβ†’0(1+x)5x2βˆ’limxβ†’0e5xx2+limxβ†’0sin(x2)x2Ξ”=limxβ†’05(1+x)42xβˆ’limxβ†’05e5x2x+1Ξ”=limxβ†’020(1+x)32βˆ’limxβ†’025e5x2+1Ξ”=10limxβ†’0(1+x)3βˆ’252limexβ†’05x+1Ξ”=10βˆ’252+1Ξ”=βˆ’32=βˆ’1.5∴limxβ†’0(1+x)5βˆ’e5x+sinx2x2=βˆ’32=βˆ’1.5

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