What is the limit as h approaches 0 of ((a^{u+h}-a^u))/h ?

The limit as h approaches 0 of ((a^{u+h}-a^u))/h is a^uln(a)

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limit as h approaches 0 of ((a^{u+h}-a^u))/h

h \to 0 lim (\frac{( a ^{u + h} - a ^{u} )}{h})

a^{u} ln (a)

Solution steps

h \to 0 lim (\frac{a ^{u + h} - a ^{u}}{h})

Apply L'Hopital's Rule

= h \to 0 lim (\frac{a ^{h + u} ln ( a )}{1})

Plug in the value

h = 0

= \frac{a ^{0 + u} ln ( a )}{1}

Simplify

= a^{u} ln (a)

\int e^{- a x} sin (b x) d x

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\frac{d}{d x} (c_{1} cos (x) + c_{2} sin (x))

y^{^{''}} - 4 y^{^{'}} + 4 y = 0

\int \frac{x}{7 x ^{2} + 3} d x

What is the limit as h approaches 0 of ((a^{u+h}-a^u))/h ?
The limit as h approaches 0 of ((a^{u+h}-a^u))/h is a^uln(a)