What is the limit as x approaches 0 of (e^x-1)/(16x) ?

The limit as x approaches 0 of (e^x-1)/(16x) is 1/16

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limit as x approaches 0 of (e^x-1)/(16x)

x \to 0 lim (\frac{e ^{x} - 1}{16 x})

\frac{1}{16}

Decimal Notation

0.0625

Solution steps

x \to 0 lim (\frac{e ^{x} - 1}{16 x})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= \frac{1}{16} \cdot x \to 0 lim (\frac{e ^{x} - 1}{x})

Apply L'Hopital's Rule

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

= \frac{1}{16} \cdot x \to 0 lim (\frac{e ^{x}}{1})

Plug in the value

x = 0

= \frac{1}{16} \cdot \frac{e ^{0}}{1}

Simplify

\frac{1}{16} \cdot \frac{e ^{0}}{1} : \frac{1}{16}

= \frac{1}{16}

\frac{d}{d x} (x^{4} - 8 x^{2} + 1)

\int arcsin (y) d y

\int_{- \infty}^{0} \frac{1}{1 + x ^{2}} d x

\frac{d}{d x} (ln (5 x + 1))

f (x) = \frac{19}{x ^{5}}

What is the limit as x approaches 0 of (e^x-1)/(16x) ?
The limit as x approaches 0 of (e^x-1)/(16x) is 1/16