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

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

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limit as h approaches 0 of (e^h-1)/(2h)

h \to 0 lim (\frac{e ^{h} - 1}{2 h})

\frac{1}{2}

Decimal Notation

0.5

Solution steps

h \to 0 lim (\frac{e ^{h} - 1}{2 h})

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

= \frac{1}{2} \cdot h \to 0 lim (\frac{e ^{h} - 1}{h})

Apply L'Hopital's Rule

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

= \frac{1}{2} \cdot h \to 0 lim (\frac{e ^{h}}{1})

Plug in the value

h = 0

= \frac{1}{2} \cdot \frac{e ^{0}}{1}

Simplify

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

= \frac{1}{2}

h \to 0 lim (cos (h))

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

x \to 0 lim (sin (x) - tan (x))

x \to 0 + lim (lo g_{e} (x))

x \to 5 + lim (f (x))

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