Limited yet powerful (II)

Calculus Level 2

lim ⁡ x → ∞ ( 1 − 1 x ) x = e N \lim_{x\rightarrow\infty} (1-\frac{1}{x})^{x} =e^{N} . N = ? N=?

This problem differs by just one stroke.


The answer is -1.

Noel Lo by Noel Lo , 24, Singapore

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2 solutions

Noel Lo
Nov 30, 2015

lim ⁡ x → ∞ ( 1 − 1 x ) x = e x p ( lim ⁡ x → ∞ l n ( 1 − 1 x ) x ) = e x p ( lim ⁡ x → ∞ l n ( x − 1 x ) x ) \lim_{x\rightarrow\infty} (1-\frac{1}{x})^{x} =exp(\lim_{x\rightarrow\infty} ln(1-\frac{1}{x})^{x})=exp(\lim_{x\rightarrow\infty} ln(\frac{x-1}{x})^{x}) = e x p ( lim ⁡ x → ∞ l n ( x − 1 ) − l n ( x ) 1 x ) = e x p ( lim ⁡ x → ∞ 1 x − 1 − 1 x − 1 x 2 ) = e x p ( lim ⁡ x → ∞ 1 x ( x − 1 ) − 1 x 2 ) = e x p ( lim ⁡ x → ∞ − x x − 1 ) =exp(\lim_{x\rightarrow\infty} \frac{ln(x-1)-ln(x)}{\frac{1}{x}})=exp(\lim_{x\rightarrow\infty} \frac{\frac{1}{x-1}-\frac{1}{x}}{\frac{-1}{x^2}})=exp(\lim_{x\rightarrow\infty} \frac{\frac{1}{x(x-1)}}{\frac{-1}{x^2}})=exp(\lim_{x\rightarrow\infty} \frac{-x}{x-1}) = e x p ( − 1 ) = e − 1 =exp(-1)=e^{-1}

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Don't understand at all. What is exp. Sorry i am stupid.

Davin Shaun - 5 years, 6 months ago

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exp(x)=e^x

Luke Smith - 10 months, 2 weeks ago
Oliver Petts
Jul 30, 2020

( 1 − 1 x ) x = ( x − 1 x ) x = ( x x − 1 ) − x = ( x x − 1 ) − x ⋅ ( x x − 1 ) 1 ⋅ ( x x − 1 ) − 1 = ( x x − 1 ) 1 − x ⋅ ( x x − 1 ) − 1 = [ ( x x − 1 ) x − 1 ] − 1 ⋅ ( x x − 1 ) − 1 \left( 1-\frac { 1 }{ x } \right) ^{ x }=\left( \frac { x-1 }{ x } \right) ^{ x }=\left( \frac { x }{ x-1 } \right) ^{ -x }=\left( \frac { x }{ x-1 } \right) ^{ -x }\cdot \left( \frac { x }{ x-1 } \right) ^{ 1 }\cdot \left( \frac { x }{ x-1 } \right) ^{ -1 }=\left( \frac { x }{ x-1 } \right) ^{ 1-x }\cdot \left( \frac { x }{ x-1 } \right) ^{ -1 }=\left[ \left( \frac { x }{ x-1 } \right) ^{ x-1 } \right] ^{ -1 }\cdot \left( \frac { x }{ x-1 } \right) ^{ -1 }

As lim ⁡ x → ∞ ( x x − 1 ) x − 1 = e \lim _{ x\rightarrow \infty }{ \left( \frac { x }{ x-1 } \right) ^{ x-1 } } =e (let y = x - 1) then we have,

[ ( x x − 1 ) x − 1 ] − 1 ⋅ ( x x − 1 ) − 1 = e − 1 ⋅ ( x x − 1 ) − 1 = e − 1 ⋅ ( x − 1 x ) = e − 1 ⋅ ( 1 − 1 x ) \left[ \left( \frac { x }{ x-1 } \right) ^{ x-1 } \right] ^{ -1 }\cdot \left( \frac { x }{ x-1 } \right) ^{ -1 }=e^{ -1 }\cdot \left( \frac { x }{ x-1 } \right) ^{ -1 }=e^{ -1 }\cdot \left( \frac { x-1 }{ x } \right) =e^{ -1 }\cdot \left( 1-\frac { 1 }{ x } \right)

So, lim ⁡ x → ∞ ( 1 − 1 x ) x = e − 1 ⋅ lim ⁡ x → ∞ ( 1 − 1 x ) = e − 1 ⋅ 1 = e − 1 \lim _{ x\rightarrow \infty }{ \left( 1-\frac { 1 }{ x } \right) ^{ x } } =e^{ -1 }\cdot \lim _{ x\rightarrow \infty }{ \left( 1-\frac { 1 }{ x } \right) } =e^{ -1 }\cdot 1=e^{ -1 }

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