What is the limit as x approaches 0 of (9^x+x)^{2/x} ?

The limit as x approaches 0 of (9^x+x)^{2/x} is e^{2(2ln(3)+1)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

limit as x approaches 0 of (9^x+x)^{2/x}

x \to 0 lim ((9^{x} + x)^{\frac{2}{x}})

e^{2 (2 l n (3) + 1)}

Decimal

598.51354 \dots

Solution steps

x \to 0 lim ((9^{x} + x)^{\frac{2}{x}})

Apply exponent rules

= x \to 0 lim (e^{\frac{2}{x} l n (9^{x} + x)})

Apply the Limit Chain Rule:

e^{2 (2 l n (3) + 1)}

= e^{2 (2 l n (3) + 1)}

x \to 3 lim (3 x^{2} + x)

x \to 1 lim ([\frac{x ^{3} - 1}{x ^{2} - 1}])

x \to 1 lim (\frac{5 x + 3 x - 2}{x ^{2} - 1})

x \to 0 lim (\frac{( 1 - cos ^{2} ( x ) )}{ln ( 1 + x ^{2} )})

x \to 0 lim (x^{9} e^{- x^{8}})

What is the limit as x approaches 0 of (9^x+x)^{2/x} ?
The limit as x approaches 0 of (9^x+x)^{2/x} is e^{2(2ln(3)+1)}