What is the limit as x approaches infinity of ((x+1}{x+2})^{\frac{-4x)/5} ?

The limit as x approaches infinity of ((x+1}{x+2})^{\frac{-4x)/5} is e^{4/5}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

limit as x approaches infinity of ((x+1}{x+2})^{\frac{-4x)/5}

x \to \infty lim ((\frac{x + 1}{x + 2})^{\frac{- 4 x}{5}})

e^{\frac{4}{5}}

Solution steps

x \to \infty lim ((\frac{x + 1}{x + 2})^{\frac{- 4 x}{5}})

Apply exponent rules

= x \to \infty lim (e^{\frac{- 4 x}{5} l n (\frac{x + 1}{x + 2})})

Apply the Limit Chain Rule:

e^{\frac{4}{5}}

= e^{\frac{4}{5}}

f (x) = \frac{x ^{3} + 7}{x}

\int \frac{3}{x ^{5}} d x

\int \frac{( 16 x ^{2} )}{( x ^{2} + 9 ) ^{5}} d x

\int \frac{x ^{2} - x + 1}{x ^{3}} d x

\frac{d}{d x} (- 5 sin (x) cos (x))

What is the limit as x approaches infinity of ((x+1}{x+2})^{\frac{-4x)/5} ?
The limit as x approaches infinity of ((x+1}{x+2})^{\frac{-4x)/5} is e^{4/5}