1 + x = ( 1 + x ) 1 / 2 \sqrt{1+x}=(1+x)^{1/2} 1 + x = ( 1 + x ) 1/2
= 1 + 1 2 x + 1 2 ( 1 2 − 1 ) 2 ! x 2 + 1 2 ( 1 2 − 1 ) ( 1 2 − 2 ) 3 ! x 3 + . . . =1+\dfrac{1}{2}x+\dfrac{\dfrac{1}{2}(\dfrac{1}{2}-1)}{2!}x^2+\dfrac{\dfrac{1}{2}(\dfrac{1}{2}-1)(\dfrac{1}{2}-2)}{3!}x^3+... = 1 + 2 1 x + 2 ! 2 1 ( 2 1 − 1 ) x 2 + 3 ! 2 1 ( 2 1 − 1 ) ( 2 1 − 2 ) x 3 + ...
f ( x ) = x 1 + x = x + 1 2 x 2 − 1 8 x 3 + 1 16 x 3 − 5 128 x 4 + . . . f(x)=x\sqrt{1+x}=x+\dfrac{1}{2}x^2-\dfrac{1}{8}x^3+\dfrac{1}{16}x^3-\dfrac{5}{128}x^4+... f ( x ) = x 1 + x = x + 2 1 x 2 − 8 1 x 3 + 16 1 x 3 − 128 5 x 4 + ... The function g ( x ) = x ( cosh x ) 1 / x g(x)=x(\cosh x)^{1/x} g ( x ) = x ( cosh x ) 1/ x is not defined at x = 0. x=0. x = 0. Therefore we cannot write down the Maclaurin expression of the following function.
By the calculator
g ( x ) = x ( cosh x ) 1 / x ≈ x + 1 2 x 2 + 1 8 x 3 − 1 16 x 4 − 5 128 x 5 + . . . g(x)=x(\cosh x)^{1/x}\approx x+\dfrac{1}{2}x^2+\dfrac{1}{8}x^3-\dfrac{1}{16}x^4-\dfrac{5}{128}x^5+... g ( x ) = x ( cosh x ) 1/ x ≈ x + 2 1 x 2 + 8 1 x 3 − 16 1 x 4 − 128 5 x 5 + ...
x 1 + x + x ( cosh x ) 1 / x ≈ 2 x + x 2 − 5 64 x 5 x\sqrt{1+x}+x(\cosh x)^{1/x}\approx2x+x^2-\dfrac{5}{64}x^5 x 1 + x + x ( cosh x ) 1/ x ≈ 2 x + x 2 − 64 5 x 5
e x − 1 = x + 1 2 x 2 + 1 6 x 3 + 1 24 x 4 + 1 120 x 5 + . . . e^x-1=x+\dfrac{1}{2}x^2+\dfrac{1}{6}x^3+\dfrac{1}{24}x^4+\dfrac{1}{120}x^5+... e x − 1 = x + 2 1 x 2 + 6 1 x 3 + 24 1 x 4 + 120 1 x 5 + ...
( e x − 1 ) 3 ≈ x 3 + 3 2 x 4 + 5 4 x 5 (e^x-1)^3\approx x^3+\dfrac{3}{2}x^4+\dfrac{5}{4}x^5 ( e x − 1 ) 3 ≈ x 3 + 2 3 x 4 + 4 5 x 5
lim x → 0 ( 4 ( x 1 + x + x ( cosh x ) 1 / x ) ( e x − 1 ) 3 ) \lim\limits_{x\to0}(\dfrac{4(x\sqrt{1+x}+x(\cosh x)^{1/x})}{(e^x-1)^3}) x → 0 lim ( ( e x − 1 ) 3 4 ( x 1 + x + x ( cosh x ) 1/ x ) )
= lim x → 0 ( 4 ( 2 x + x 2 − 5 64 x 5 ) x 3 + 3 2 x 4 + 5 4 x 5 ) =\lim\limits_{x\to0}(\dfrac{4(2x+x^2-\dfrac{5}{64}x^5)}{x^3+\dfrac{3}{2}x^4+\dfrac{5}{4}x^5}) = x → 0 lim ( x 3 + 2 3 x 4 + 4 5 x 5 4 ( 2 x + x 2 − 64 5 x 5 ) )
= − 1 4 , x ≠ 0 =-\dfrac{1}{4},x\not=0 = − 4 1 , x = 0
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