What is the limit as x approaches 4 of (x^2-16)/(2x^2+7x-4) ?

The limit as x approaches 4 of (x^2-16)/(2x^2+7x-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 x approaches 4 of (x^2-16)/(2x^2+7x-4)

x \to 4 lim (\frac{x ^{2} - 16}{2 x ^{2} + 7 x - 4})

0

Solution steps

x \to 4 lim (\frac{x ^{2} - 16}{2 x ^{2} + 7 x - 4})

Plug in the value

x = 4

= \frac{4 ^{2} - 16}{2 \cdot 4 ^{2} + 7 \cdot 4 - 4}

Simplify

\frac{4 ^{2} - 16}{2 \cdot 4 ^{2} + 7 \cdot 4 - 4} : 0

= 0

sin (2 x) d x + cos (5 y) d y = 0

\frac{d}{d x} (ln (x^{2}) + y^{2} - 1)

\frac{d}{d x} ((\frac{( 3 x + 1 )}{x ^{2}})^{3})

x \to + 0 lim (x (x - [x]))

\frac{\partial}{\partial x} (- 2 y^{3} sin (2 x))

What is the limit as x approaches 4 of (x^2-16)/(2x^2+7x-4) ?
The limit as x approaches 4 of (x^2-16)/(2x^2+7x-4) is 0