What is the limit as x approaches 0 of (2^x+x)^{1/x} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

limit as x approaches 0 of (2^x+x)^{1/x}

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

2 e

Decimal

5.43656 \dots

Solution steps

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

Apply exponent rules

= x \to 0 lim (e^{\frac{1}{x} l n (2^{x} + x)})

Apply the Limit Chain Rule:

2 e

= 2 e

x \to 0 + lim (ln ((x + 1)^{x}))

x \to 1 + lim (ln (x) ln (x - 1))

x \to 4 lim (3 (x - 4)^{x - 4})

x \to 0 lim (2 x csc (x))

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

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