What is the limit as h approaches 0 of ((2+h)^5-32)/h ?

The limit as h approaches 0 of ((2+h)^5-32)/h is 80

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limit as h approaches 0 of ((2+h)^5-32)/h

h \to 0 lim (\frac{( 2 + h ) ^{5} - 32}{h})

80

Solution steps

h \to 0 lim (\frac{( 2 + h ) ^{5} - 32}{h})

Simplify

\frac{( 2 + h ) ^{5} - 32}{h} : 8 h + (h + 2)^{4} + 2 (h + 2)^{3} + 4 (h + 2)^{2} + 32

= h \to 0 lim (8 h + (h + 2)^{4} + 2 (h + 2)^{3} + 4 (h + 2)^{2} + 32)

Plug in the value

h = 0

= 8 \cdot 0 + (0 + 2)^{4} + 2 (0 + 2)^{3} + 4 (0 + 2)^{2} + 32

Simplify

8 \cdot 0 + (0 + 2)^{4} + 2 (0 + 2)^{3} + 4 (0 + 2)^{2} + 32 : 80

= 80

x \to - \infty lim (x^{2 n})

x \to \infty lim (\frac{2}{+ x})

x \to 1 + lim (\frac{x ^{3} - 1}{x - 1})

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

x \to \infty lim (\frac{e ^{1 - x}}{1 + x})

What is the limit as h approaches 0 of ((2+h)^5-32)/h ?
The limit as h approaches 0 of ((2+h)^5-32)/h is 80