What is the limit as x approaches 5 of (x+3)/(x^2-9) ?

The limit as x approaches 5 of (x+3)/(x^2-9) is 1/2

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limit as x approaches 5 of (x+3)/(x^2-9)

x \to 5 lim (\frac{x + 3}{x ^{2} - 9})

\frac{1}{2}

Decimal Notation

0.5

Solution steps

x \to 5 lim (\frac{x + 3}{x ^{2} - 9})

Plug in the value

x = 5

= \frac{5 + 3}{5 ^{2} - 9}

Simplify

\frac{5 + 3}{5 ^{2} - 9} : \frac{1}{2}

= \frac{1}{2}

x \to 0 lim (\frac{2}{x ( x - 2 )})

x \to 0 lim (\frac{7 ^{x} - 8 ^{x}}{x})

x \to π lim (\frac{x}{6} - \frac{π}{2})

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

What is the limit as x approaches 5 of (x+3)/(x^2-9) ?
The limit as x approaches 5 of (x+3)/(x^2-9) is 1/2