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

The limit as x approaches 1 of (x-3)/(x^2-9) is 1/4

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

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

\frac{1}{4}

Decimal Notation

0.25

Solution steps

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

Plug in the value

x = 1

= \frac{1 - 3}{1 ^{2} - 9}

Simplify

\frac{1 - 3}{1 ^{2} - 9} : \frac{1}{4}

= \frac{1}{4}

\frac{d}{d x} (lo g_{4} (x^{2} - 9 x))

\int \frac{3}{- 1 - x} d x

x \to 0 lim (\frac{1}{2} + x^{2})

\int 9 x^{2} sin (π x) d x

\frac{d}{d t} (3 t^{3})

What is the limit as x approaches 1 of (x-3)/(x^2-9) ?
The limit as x approaches 1 of (x-3)/(x^2-9) is 1/4