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

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

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

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

2

Solution steps

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

\int_{0}^{1} \frac{x ^{4} + 9}{x} d x

\frac{6}{s} + \frac{1}{s ^{2} + 16}

\frac{d}{d x} (a (e^{b x} + b e^{b x} x))

\frac{\partial}{\partial x} (x (4 - x^{2}))

y^{^{''}} - 4 y^{^{'}} + 4 y = x^{2} e^{x}

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