What is the limit as h approaches 0 of ((1+h)^9-1)/h ?

The limit as h approaches 0 of ((1+h)^9-1)/h is 9

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limit as h approaches 0 of ((1+h)^9-1)/h

h \to 0 lim (\frac{( 1 + h ) ^{9} - 1}{h})

9

Solution steps

h \to 0 lim (\frac{( 1 + h ) ^{9} - 1}{h})

Simplify

\frac{( 1 + h ) ^{9} - 1}{h} : (h^{2} + 3 h + 3) ((h + 1)^{6} + (h + 1)^{3} + 1)

= h \to 0 lim ((h^{2} + 3 h + 3) ((h + 1)^{6} + (h + 1)^{3} + 1))

Plug in the value

h = 0

= (0^{2} + 3 \cdot 0 + 3) ((0 + 1)^{6} + (0 + 1)^{3} + 1)

Simplify

(0^{2} + 3 \cdot 0 + 3) ((0 + 1)^{6} + (0 + 1)^{3} + 1) : 9

= 9

\frac{d}{d x} (x^{3} + 6 x^{2})

5 x y (\frac{d y}{d x}) = 3

\int_{0}^{2} x 1 + 2 x^{2} d x

\int 3 + tan^{2} (x) d x

3 csc (x)

What is the limit as h approaches 0 of ((1+h)^9-1)/h ?
The limit as h approaches 0 of ((1+h)^9-1)/h is 9