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

The limit as x approaches-1-of (3+x)/(x^2-1) is infinity

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

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

\infty

Solution steps

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

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

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

x \to a lim [f (x) \cdot g (x)] = x \to a lim f (x) \cdot x \to a lim g (x)

With the exception of indeterminate form

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

x \to - 1 - lim (3 + x) = 2

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

= 2 \cdot \infty

Apply Infinity Property:

c \cdot \infty = \infty

= \infty

\frac{x + x cos ( x ) - 2 sin ( x )}{x ^{3}}

\frac{d}{d x} ((2 x - 5)^{4} (8 x^{2} - 5)^{- 3})

\frac{\partial}{\partial u} (\frac{uv}{v + 1})

\frac{d}{d x} (x^{3} - 6 x^{2} + 9 x - 4)

\int (x + 1) (x^{2} + 2 x) d x

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