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

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

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

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

2

Solution steps

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

Apply L'Hopital's Rule

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

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

Plug in the value

h = 0

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

Simplify

= 2

\frac{d}{d x} (x \cdot cos (x))

f (x) = 2 x 5 - 2 x, at x = - 1

\int r d r

y^{^{'}} = 1 + y^{2}

f (x) = 3 x^{2} - x + 3

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