What is the limit as x approaches 3 of 1/((x^2-9)) ?

The limit as x approaches 3 of 1/((x^2-9)) is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

limit as x approaches 3 of 1/((x^2-9))

x \to 3 lim (\frac{1}{( x ^{2} - 9 )})

diverges

Solution steps

x \to 3 lim (\frac{1}{x ^{2} - 9})

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

then the limit does not exist

x \to 3 + lim (\frac{1}{x ^{2} - 9}) = \infty

x \to 3 - lim (\frac{1}{x ^{2} - 9}) = - \infty

= diverges

\int 2 x e^{- 2 x^{2}} d x

f (x) = x^{5} \cdot e^{x}

n = 1 \sum \infty \frac{( x + 3 ) ^{n}}{n}

\int 20 + 0.6 x d x

(\frac{e}{16})^{x}

What is the limit as x approaches 3 of 1/((x^2-9)) ?
The limit as x approaches 3 of 1/((x^2-9)) is diverges