What is the limit as x approaches 0 of (ln(e^x+x))/x ?

The limit as x approaches 0 of (ln(e^x+x))/x is 2

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limit as x approaches 0 of (ln(e^x+x))/x

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

2

Solution steps

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

= \frac{\frac{e ^{0} + 1}{e ^{0} + 0}}{1}

Simplify

\frac{\frac{e ^{0} + 1}{e ^{0} + 0}}{1} : 2

= 2

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

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

x \to 0 + lim (\frac{sin ( 3 x )}{x})

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

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

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