What is the limit as x approaches infinity of (e^{2x-7})/(6x) ?

The limit as x approaches infinity of (e^{2x-7})/(6x) is infinity

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limit as x approaches infinity of (e^{2x-7})/(6x)

x \to \infty lim (\frac{e ^{2 x - 7}}{6 x})

\infty

Solution steps

x \to \infty lim (\frac{e ^{2 x - 7}}{6 x})

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

= \frac{1}{6} \cdot x \to \infty lim (\frac{e ^{2 x - 7}}{x})

Simplify

e^{2 x - 7} : e^{2 x} e^{- 7}

= \frac{1}{6} \cdot x \to \infty lim (\frac{e ^{2 x} e ^{- 7}}{x})

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

= \frac{1}{6} e^{- 7} \cdot x \to \infty lim (\frac{e ^{2 x}}{x})

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{6} \cdot \frac{1}{e ^{7}} \cdot x \to \infty lim (\frac{e ^{2 x}}{x})

Apply L'Hopital's Rule

= \frac{1}{6} \cdot \frac{1}{e ^{7}} \cdot x \to \infty lim (\frac{e ^{2 x} \cdot 2}{1})

Simplify

= \frac{1}{6} \cdot \frac{1}{e ^{7}} \cdot x \to \infty lim (2 e^{2 x})

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

= \frac{1}{6} \cdot \frac{1}{e ^{7}} \cdot 2 \cdot x \to \infty lim (e^{2 x})

Apply the Limit Chain Rule:

\infty

= \frac{1}{6} \cdot \frac{1}{e ^{7}} \cdot 2 \cdot \infty

Apply Infinity Property:

c \cdot \infty = \infty

= \infty

What is the limit as x approaches infinity of (e^{2x-7})/(6x) ?
The limit as x approaches infinity of (e^{2x-7})/(6x) is infinity