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

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

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

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

\infty

Solution steps

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

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

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

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

= 5 \cdot x \to 3 + lim (x \frac{1}{x ^{2} - 9})

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

= 5 \cdot x \to 3 + lim (x) \cdot x \to 3 + lim (\frac{1}{x ^{2} - 9})

x \to 3 + lim (x) = 3

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

= 5 \cdot 3 \cdot \infty

Apply Infinity Property:

c \cdot \infty = \infty

= \infty

\frac{d}{d x} (ln (cosh (2 x)))

2 y^{^{''}} + y^{^{'}} - 10 y = 0

\int 3 e^{- 0.8 x} d x

y = 4 x, y = 9 x^{2}

y = 6 e^{x} + \frac{4}{3 x}

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