炘涛 發問時間: 科學數學 · 5 年前

請問下列極限怎麼算?

1. lim x^x

x->0

2. lim x^(2x)

x-->0

3. lim {1/x-1/[(e^x)-1]}

x->0

4. lim (1-1/x)^x

x->無限

5. lim (1-x)^(1/x)

x->0

6. show that lim(x->無限) (x^r)/e^x=0 for all x屬於R

1 個解答

評分
  • 匿名使用者
    5 年前
    最佳解答

    1.

    lim(x→0) xˣ

    = e^[ lim(x→0) ln(xˣ) ]

    = e^[ lim(x→0⁺) x ln x ]

    = e^[ lim(x→0⁺) (ln x) / (1/x) ] ...... ( -∞/∞ )

    = e^[ lim(x→0⁺) (1/x) / (-1/x²) ] ...... ( 利用羅必達法則 )

    = e^[ - lim(x→0⁺) x ]

    = 1

    🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘

    2.

    lim(x→0) x²ˣ

    = e^[ lim(x→0) ln(x²ˣ) ]

    = e^[ lim(x→0⁺) 2x ln x ]

    = e^[ 2 lim(x→0⁺) (ln x) / (1/x) ] ...... ( -∞/∞ )

    = e^[ 2 lim(x→0⁺) (1/x) / (-1/x²) ] ...... ( 利用羅必達法則 )

    = e^[ -2 lim(x→0⁺) x ]

    = 1

    🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘

    3.

    lim(x→0) 1/x - 1/(eˣ - 1)

    = lim(x→0) (eˣ - 1 - x)/[ x(eˣ - 1) ]

    = lim(x→0) (eˣ - x - 1)/(xeˣ - x) ...... ( 0/0 )

    = lim(x→0) (eˣ - 1 - 0)/(eˣ + xeˣ - 1) ...... ( 利用羅必達法則 )

    = lim(x→0) (eˣ - 1)/(eˣ + xeˣ - 1) ...... ( 0/0 )

    = lim(x→0) (eˣ - 0)/(eˣ + eˣ + xeˣ - 0) ...... ( 利用羅必達法則 )

    = lim(x→0) 1/(2 + x)

    = 1/2

    🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘

    4.

    lim(x→∞) (1 - 1/x)ˣ

    = e^[ lim(x→∞) ln (1 - 1/x)ˣ ]

    = e^[ lim(x→∞) ln (1 - 1/x) / (1/x) ] ...... ( 0/0 )

    = e^[ lim(x→∞) 1/(1 - 1/x)*(1/x²) / (-1/x²) ] ...... ( 利用羅必達法則 )

    = e^[ - lim(x→∞) 1/(1 - 1/x) ]

    = 1/e

    🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘

    5.

    lim(x→0) (1 - x)^(1/x)

    = e^[ lim(x→0) ln (1 - x)^(1/x) ]

    = e^[ lim(x→0) ln (1 - x) / x ] ...... ( 0/0 )

    = e^[ lim(x→0) -1/(1 - x) / 1 ] ...... ( 利用羅必達法則 )

    = e^[ - lim(x→0) 1/(1 - x) ]

    = 1/e

    🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘🔘

    6.

    lim(x→∞) xʳ/eˣ ...... [ 假設 r 為實數 ]

    若 r < 0,

    lim(x→∞) xʳ/eˣ

    = lim(x→∞) 1/(eˣ x⁻ʳ)

    = 0

    若 r = 0,

    lim(x→∞) xʳ/eˣ

    = lim(x→∞) 1/eˣ

    = 0

    若 r > 0,

    lim(x→∞) xʳ / eˣ ...... ( ∞/∞ )

    = lim(x→∞) rxʳ⁻¹ / eˣ ...... ( 利用羅必達法則 )

    = r lim(x→∞) xʳ⁻¹ / eˣ ...... ( ∞/∞ )

    = r lim(x→∞) (r - 1)xʳ⁻² / eˣ ...... ( 利用羅必達法則 )

    ...

    = r(r - 1)(r - 2)...(r - n + 1) lim(x→∞) xʳ⁻ⁿ / eˣ ...... ( 利用羅必達法則 )

    當 r - n + 1 > 0 及 r - n < 0 時,

    = r(r - 1)(r - 2)...(r - n + 1) lim(x→∞) 1/(eˣ xⁿ⁻ʳ)

    = 0

    ∴ lim(x→∞) xʳ/eˣ = 0

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