What is the limit as y approaches 2 of (y^3-6y^2+12y-8)/(y^4-4y^3+16y-16) ?

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

limit as y approaches 2 of (y^3-6y^2+12y-8)/(y^4-4y^3+16y-16)

y \to 2 lim (\frac{y ^{3} - 6 y ^{2} + 12 y - 8}{y ^{4} - 4 y ^{3} + 16 y - 16})

\frac{1}{4}

Decimal

0.25

Solution steps

y \to 2 lim (\frac{y ^{3} - 6 y ^{2} + 12 y - 8}{y ^{4} - 4 y ^{3} + 16 y - 16})

Simplify

\frac{y ^{3} - 6 y ^{2} + 12 y - 8}{y ^{4} - 4 y ^{3} + 16 y - 16} : \frac{1}{y + 2}

= y \to 2 lim (\frac{1}{y + 2})

Plug in the value

y = 2

= \frac{1}{2 + 2}

Simplify

= \frac{1}{4}

What is the limit as y approaches 2 of (y^3-6y^2+12y-8)/(y^4-4y^3+16y-16) ?
The limit as y approaches 2 of (y^3-6y^2+12y-8)/(y^4-4y^3+16y-16) is 1/4