What is the limit as x approaches-1 of ((x-1))/(x+1) ?

The limit as x approaches-1 of ((x-1))/(x+1) 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-1 of ((x-1))/(x+1)

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

diverges

Solution steps

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

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

then the limit does not exist

x \to - 1 + lim (\frac{x - 1}{x + 1}) = - \infty

x \to - 1 - lim (\frac{x - 1}{x + 1}) = \infty

= diverges

x \to \frac{5}{6} lim (\frac{1}{6 x - 5})

x \to 8 lim (\frac{x - 8}{x + 9})

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

u \to 3.3 lim (u sec (u))

x \to 0 lim (\frac{sin ( 3 x )}{e ^{2 x} - 1})

What is the limit as x approaches-1 of ((x-1))/(x+1) ?
The limit as x approaches-1 of ((x-1))/(x+1) is diverges