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
- 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
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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
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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
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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
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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
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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