lim-0-t-dt-

Question Number 1573 by 123456 last updated on 20/Aug/15

\lim_{ϵ \to + \infty} \underset{0}{\int^{ϵ}} ϵ^{- t} d t = ?

Commented by 112358 last updated on 21/Aug/15

L e t I (ϵ) = \int_{0}^{ϵ} ϵ^{- t} d t (ϵ > 0)

I f u = - t \Rightarrow - d u = d t

∴ I (ϵ) = - \int_{0}^{- ϵ} ϵ^{u} d u = - \frac{ϵ^{u}}{l n ϵ} ∣_{0}^{- ϵ}

I (ϵ) = - \frac{1}{l n ϵ} (ϵ^{- ϵ} - ϵ^{0})

= \frac{1}{l n ϵ} (1 - \frac{1}{ϵ^{ϵ}})

I (ϵ) = \frac{ϵ^{ϵ} - 1}{l n ϵ^{ϵ^{ϵ}}}

∴ I f L = \lim_{ϵ \to + \infty} I (ϵ) = \lim_{ϵ \to + \infty} \frac{1}{l n ϵ} (1 - \frac{1}{ϵ^{ϵ}})

L = (\lim_{ϵ \to + \infty} \frac{1}{l n ϵ}) (\underset{ϵ \to + \infty}{\lim 1} - \lim_{ϵ \to + \infty} \frac{1}{ϵ^{ϵ}})

L = (\frac{\underset{ϵ \to + \infty}{\lim 1}}{\lim_{ϵ \to + \infty} l n ϵ}) (\underset{ϵ \to + \infty}{\lim 1} - \frac{\underset{ϵ \to + \infty}{\lim 1}}{\lim_{ϵ \to + \infty} ϵ^{ϵ}})

∵ \underset{ϵ \to + \infty}{\lim 1} = 1, \lim_{ϵ \to + \infty} l n ϵ = + \infty, \lim_{ϵ \to + \infty} ϵ^{ϵ} = + \infty

\Rightarrow \frac{\underset{ϵ \to + \infty}{\lim 1}}{\lim_{ϵ \to + \infty} l n ϵ} = \frac{1}{+ \infty} = 0

\underset{ϵ \to + \infty}{\lim 1} - \frac{\underset{ϵ \to + \infty}{\lim 1}}{\lim_{ϵ \to + \infty} ϵ^{ϵ}} = 1 - \frac{1}{+ \infty} = 1

∴ L = 0 \times 1 = 0

P r o o f o f L = \lim_{x \to + \infty} x^{x} = \infty .

I n f o r m a l l y,

L = \lim_{x \to + \infty} x^{x} = \lim_{x \to + \infty} e^{{l n x}^{x}}

L = \lim_{x \to + \infty} e^{x l n x} = e^{\lim_{x \to + \infty} x l n x}

L = e x p ((\lim_{x \to 0} x) (\lim_{x \to + \infty} l n x))

\lim_{x \to + \infty} l n x = \infty, \lim_{x \to + \infty} x = \infty

∴ L = e^{\infty \times \infty} = \infty

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