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

The limit as x approaches-1^{-1} of x/(x+1) is diverges

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limit as x approaches-1^{-1} of x/(x+1)

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

diverges

Solution steps

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

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

then the limit does not exist

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

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

= diverges

\int \frac{9}{x + 9 + x} d x

d er i v a t i v e f (x) = x^{2} (1 - 8 x)

t an g e n t ln (2 x + 1)

f^{^{''}} (x) = \frac{x ^{2} - 2 x + 1}{x ^{2}}

\frac{d}{d x} (x - x^{3} sin (x))

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