What is the limit as x approaches infinity of ((e^{(15)/x}-1))/((15)/x) ?

The limit as x approaches infinity of ((e^{(15)/x}-1))/((15)/x) is 1

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limit as x approaches infinity of ((e^{(15)/x}-1))/((15)/x)

x \to \infty lim \frac{( e ^{\frac{15}{x}} - 1 )}{\frac{15}{x}}

1

Solution steps

x \to \infty lim (\frac{e ^{\frac{15}{x}} - 1}{\frac{15}{x}})

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

= \frac{1}{15} \cdot x \to \infty lim (\frac{e ^{\frac{15}{x}} - 1}{\frac{1}{x}})

Simplify

= \frac{1}{15} \cdot x \to \infty lim (x (e^{\frac{15}{x}} - 1))

Rewrite for L'Hopital

= \frac{1}{15} \cdot x \to \infty lim (\frac{- 1 + e ^{\frac{15}{x}}}{\frac{1}{x}})

Apply L'Hopital's Rule

: x \to \infty lim \frac{- \frac{15 e ^{\frac{15}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}}

= \frac{1}{15} \cdot x \to \infty lim \frac{- \frac{15 e ^{\frac{15}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}}

Simplify

\frac{- \frac{15 e ^{\frac{15}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}} : 15 e^{\frac{15}{x}}

= \frac{1}{15} \cdot x \to \infty lim (15 e^{\frac{15}{x}})

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

= \frac{1}{15} \cdot 15 \cdot x \to \infty lim (e^{\frac{15}{x}})

Apply the Limit Chain Rule:

1

= \frac{1}{15} \cdot 15 \cdot 1

Simplify

\frac{1}{15} \cdot 15 \cdot 1 : 1

= 1

(s + \frac{3}{2})^{2}

\int e^{\frac{- x}{k}} \cdot \frac{- x}{k} d x

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

f (x) = \frac{2 x + 5}{x + 2} \cdot 2 x - 5

\frac{d y}{d x} = x y + y^{2}

What is the limit as x approaches infinity of ((e^{(15)/x}-1))/((15)/x) ?
The limit as x approaches infinity of ((e^{(15)/x}-1))/((15)/x) is 1