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

The limit as x approaches 4 of 1/(16-x^2) is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

diverges

Solution steps

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

x \to a - lim f (x) \neq = x \to a + lim f (x)

then the limit does not exist

x \to 4 + lim (\frac{1}{16 - x ^{2}}) = - \infty

x \to 4 - lim (\frac{1}{16 - x ^{2}}) = \infty

= diverges

x \to \infty lim (\frac{x ^{\frac{3}{2}}}{( x + 1 ) ^{\frac{3}{2}}})

x \to 0 lim (sin (x) cot (2 x))

x \to 3 lim (x^{2} - 3 x + 7)

x \to 3 + lim (\frac{1}{x ^{2} - 3 x})

x \to 5 + lim (\frac{- 3}{x - 5})

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