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

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

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

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

2

Solution steps

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

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

= 2

x \to 0 + lim (\frac{1}{4 x} - 5)

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

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

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

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

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