What is the limit as h approaches 0 of (5e^x-5e^{x+h})/(3h) ?

The limit as h approaches 0 of (5e^x-5e^{x+h})/(3h) is -5/3 e^x

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limit as h approaches 0 of (5e^x-5e^{x+h})/(3h)

h \to 0 lim (\frac{5 e ^{x} - 5 e ^{x + h}}{3 h})

- \frac{5}{3} e^{x}

Solution steps

h \to 0 lim (\frac{5 e ^{x} - 5 e ^{x + h}}{3 h})

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

= \frac{1}{3} \cdot h \to 0 lim (\frac{5 e ^{x} - 5 e ^{x + h}}{h})

Apply L'Hopital's Rule

: h \to 0 lim (\frac{- 5 e ^{h + x}}{1})

= \frac{1}{3} \cdot h \to 0 lim (\frac{- 5 e ^{h + x}}{1})

Plug in the value

h = 0

= \frac{1}{3} \cdot \frac{- 5 e ^{0 + x}}{1}

Simplify

\frac{1}{3} \cdot \frac{- 5 e ^{0 + x}}{1} : - \frac{5}{3} e^{x}

= - \frac{5}{3} e^{x}

\frac{e ^{- s}}{( s ^{2} + 8 )}

\frac{d}{d x} (\frac{3 x - 1}{2 x + 1})

y^{^{'}} + 4 x^{- 1} y = x^{2}, y (1) = - 2

n = 1 \sum \infty \frac{5 ^{n}}{n ^{5}}

\frac{1}{x + 2}, 0

What is the limit as h approaches 0 of (5e^x-5e^{x+h})/(3h) ?
The limit as h approaches 0 of (5e^x-5e^{x+h})/(3h) is -5/3 e^x