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Question Number 131789 by bemath last updated on 08/Feb/21
Let ϕ = lim_(x→0)  ((x^4 +3(a^2 −(√(a^4 +x^4 )) ))/x^8 ) ; a>0  If ϕ is finite then   (a) a=(3/2)    (b) a=(√(3/2))      (c) ϕ=(1/3)    (d) ϕ=(1/9)
Letφ=limx0x4+3(a2a4+x4)x8;a>0Ifφisfinitethen(a)a=32(b)a=32(c)φ=13(d)φ=19
Answered by Dwaipayan Shikari last updated on 08/Feb/21
lim_(x→0) ((x^4 +3(a^2 −a^2 (1+(x^4 /(2a^4 ))−(x^8 /(8a^8 )))))/x^8 )=((x^4 −((3x^4 )/(2a^2 ))+((3x^8 )/(8a^6 )))/x^8 ) =ϕ  ⇒x^4 −((3x^4 )/(2a^2 ))=0  ⇒1=(3/(2a^2 ))⇒a=±(√(3/2))  ϕ=(3/(8a^6 ))=(3/8).(8/(27))=(1/9)
limx0x4+3(a2a2(1+x42a4x88a8))x8=x43x42a2+3x88a6x8=φx43x42a2=01=32a2a=±32φ=38a6=38.827=19
Answered by EDWIN88 last updated on 08/Feb/21
ϕ = lim_(x→0) (((x^4 +3a^2 )−3(√(a^4 +x^4 )))/x^8 ) =  ϕ = lim_(x→0)  (((x^4 +3a^2 )^2 −9(a^4 +x^4 ))/(x^8 ((x^4 +3a^2 )+3(√(a^4 +x^4 ))))))=  ϕ = (1/(6a^2 )) ×lim_(x→0) ((x^8 +6a^2 x^4 +9a^4 −9a^4 −9x^4 )/x^8 )  ϕ=(1/(6a^2 )) × lim_(x→0)  ((x^8 +x^4 (6a^2 −9))/x^8 )   since the value of ϕ is finite it follows that   6a^2 −9 must be is zero ; a = (√(3/2))  then ϕ = (1/(6((9/6)))) = (1/9)
φ=limx0(x4+3a2)3a4+x4x8=φ=limx0(x4+3a2)29(a4+x4)x8((x4+3a2)+3a4+x4))=φ=16a2×limx0x8+6a2x4+9a49a49x4x8φ=16a2×limx0x8+x4(6a29)x8sincethevalueofφisfiniteitfollowsthat6a29mustbeiszero;a=32thenφ=16(96)=19
Commented by liberty last updated on 08/Feb/21
Hallo Herr wie geht es dir
HalloHerrwiegehtesdir
Commented by EDWIN88 last updated on 08/Feb/21
  Hallo Herr auch.  Gute Nachrichten
HalloHerrauch.GuteNachrichten

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