What is the limit as x approaches 0 of (x+e^x-1)/x ?

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

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limit as x approaches 0 of (x+e^x-1)/x

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

2

Solution steps

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

Apply L'Hopital's Rule

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

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

Plug in the value

x = 0

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

Simplify

= 2

t \to 0 lim (\frac{1}{t - 3})

x \to - 2 lim (x - 4)

x \to 1 lim (f (x) (5 - x))

x \to 0 lim (\frac{sin ( k ) t}{t})

x \to 0 lim (\frac{sin ( x )}{8 x + tan ( x )})

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