What is the limit as y approaches 1/2 of (16y^2-4)^4 ?

The limit as y approaches 1/2 of (16y^2-4)^4 is 0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

limit as y approaches 1/2 of (16y^2-4)^4

y \to \frac{1}{2} lim ((16 y^{2} - 4)^{4})

0

Solution steps

y \to \frac{1}{2} lim ((16 y^{2} - 4)^{4})

Plug in the value

y = \frac{1}{2}

= (16 (\frac{1}{2})^{2} - 4)^{4}

Simplify

(16 (\frac{1}{2})^{2} - 4)^{4} : 0

= 0

x \to o lim (x^{x})

\frac{d}{d y} (2 x y^{4} e^{y})

f (x) 6 x^{4} - 10 x^{3} - 5 x^{2} + 16 x - 45

f (x) = \frac{2}{3 x + 1}, at x = - 1

x \to \frac{π}{3} lim (7 x cos (x))

What is the limit as y approaches 1/2 of (16y^2-4)^4 ?
The limit as y approaches 1/2 of (16y^2-4)^4 is 0