What is the limit as h approaches 0 of ((3+h)^4-81)/h ?

The limit as h approaches 0 of ((3+h)^4-81)/h is 108

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limit as h approaches 0 of ((3+h)^4-81)/h

h \to 0 lim (\frac{( 3 + h ) ^{4} - 81}{h})

108

Solution steps

h \to 0 lim (\frac{( 3 + h ) ^{4} - 81}{h})

Simplify

\frac{( 3 + h ) ^{4} - 81}{h} : (h^{2} + 6 h + 18) (h + 6)

= h \to 0 lim ((h^{2} + 6 h + 18) (h + 6))

Plug in the value

h = 0

= (0^{2} + 6 \cdot 0 + 18) (0 + 6)

Simplify

(0^{2} + 6 \cdot 0 + 18) (0 + 6) : 108

= 108

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What is the limit as h approaches 0 of ((3+h)^4-81)/h ?
The limit as h approaches 0 of ((3+h)^4-81)/h is 108