What is the limit as h approaches 0 of ((2(3+h)^4-162))/h ?

The limit as h approaches 0 of ((2(3+h)^4-162))/h is 216

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

limit as h approaches 0 of ((2(3+h)^4-162))/h

h \to 0 lim \frac{( 2 ( 3 + h ) ^{4} - 162 )}{h}

216

Solution steps

h \to 0 lim (\frac{2 ( 3 + h ) ^{4} - 162}{h})

Simplify

\frac{2 ( 3 + h ) ^{4} - 162}{h} : 2 (h^{3} + 12 h^{2} + 54 h + 108)

= h \to 0 lim (2 (h^{3} + 12 h^{2} + 54 h + 108))

Plug in the value

h = 0

= 2 (0^{3} + 12 \cdot 0^{2} + 54 \cdot 0 + 108)

Simplify

= 216

\frac{d}{d x} (3 ln (7 x))

\int - e^{3 t} d t

\frac{d}{d x} (6 x - tan (6 x))

e^{4 t} cos (2 t)

\frac{d y}{d x} = 5 x e^{y}

What is the limit as h approaches 0 of ((2(3+h)^4-162))/h ?
The limit as h approaches 0 of ((2(3+h)^4-162))/h is 216