What is the limit as x approaches infinity of (e^{5x-3})/(ln(x-2)) ?

The limit as x approaches infinity of (e^{5x-3})/(ln(x-2)) is infinity

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limit as x approaches infinity of (e^{5x-3})/(ln(x-2))

x \to \infty lim (\frac{e ^{5 x - 3}}{ln ( x - 2 )})

\infty

Solution steps

x \to \infty lim (\frac{e ^{5 x - 3}}{ln ( x - 2 )})

Apply L'Hopital's Rule

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

Simplify

= x \to \infty lim (5 e^{5 x - 3} (x - 2))

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

= 5 \cdot x \to \infty lim (e^{5 x - 3} (x - 2))

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 \infty lim (e^{5 x - 3}) \cdot x \to \infty lim (x - 2)

x \to \infty lim (e^{5 x - 3}) = \infty

x \to \infty lim (x - 2) = \infty

= 5 \cdot \infty \cdot \infty

Apply Infinity Property:

5\infty \infty = \infty

= \infty

What is the limit as x approaches infinity of (e^{5x-3})/(ln(x-2)) ?
The limit as x approaches infinity of (e^{5x-3})/(ln(x-2)) is infinity