What is the limit as x approaches infinity of 8xe^{1/x}-8x ?

The limit as x approaches infinity of 8xe^{1/x}-8x is 8

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limit as x approaches infinity of 8xe^{1/x}-8x

x \to \infty lim (8 x e^{\frac{1}{x}} - 8 x)

8

Solution steps

x \to \infty lim (8 x e^{\frac{1}{x}} - 8 x)

Apply the following algebraic property

: a + b = a (1 + \frac{b}{a})

8 x e^{\frac{1}{x}} - 8 x = x (8 e^{\frac{1}{x}} - 8)

= x \to \infty lim (x (8 e^{\frac{1}{x}} - 8))

Rewrite for L'Hopital

= x \to \infty lim (\frac{- 8 + 8 e ^{\frac{1}{x}}}{\frac{1}{x}})

Apply L'Hopital's Rule

: x \to \infty lim \frac{- \frac{8 e ^{\frac{1}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}}

= x \to \infty lim \frac{- \frac{8 e ^{\frac{1}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}}

Simplify

\frac{- \frac{8 e ^{\frac{1}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}} : 8 e^{\frac{1}{x}}

= x \to \infty lim (8 e^{\frac{1}{x}})

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

= 8 \cdot x \to \infty lim (e^{\frac{1}{x}})

Apply the Limit Chain Rule:

1

= 8 \cdot 1

Simplify

= 8

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What is the limit as x approaches infinity of 8xe^{1/x}-8x ?
The limit as x approaches infinity of 8xe^{1/x}-8x is 8