What is the limit as x approaches infinity of 4xe^{(1/x)}-4x ?

The limit as x approaches infinity of 4xe^{(1/x)}-4x is 4

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limit as x approaches infinity of 4xe^{(1/x)}-4x

x \to \infty lim (4 x e^{(\frac{1}{x})} - 4 x)

4

Solution steps

x \to \infty lim (4 x e^{\frac{1}{x}} - 4 x)

Apply the following algebraic property

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

4 x e^{\frac{1}{x}} - 4 x = x (4 e^{\frac{1}{x}} - 4)

= x \to \infty lim (x (4 e^{\frac{1}{x}} - 4))

Rewrite for L'Hopital

= x \to \infty lim (\frac{- 4 + 4 e ^{\frac{1}{x}}}{\frac{1}{x}})

Apply L'Hopital's Rule

: x \to \infty lim \frac{- \frac{4 e ^{\frac{1}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}}

= x \to \infty lim \frac{- \frac{4 e ^{\frac{1}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}}

Simplify

\frac{- \frac{4 e ^{\frac{1}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}} : 4 e^{\frac{1}{x}}

= x \to \infty lim (4 e^{\frac{1}{x}})

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

= 4 \cdot x \to \infty lim (e^{\frac{1}{x}})

Apply the Limit Chain Rule:

1

= 4 \cdot 1

Simplify

= 4

\int (2 y - \frac{1}{2 y}) d y

2 x^{2} - x^{4}

x \to - 3 lim (\frac{7}{( x + 3 ) ^{2}})

f (x) = \frac{3}{x}, at x = 2

\frac{\partial}{\partial x} (ln (7 x^{2} + 4 y^{2} + 8))

What is the limit as x approaches infinity of 4xe^{(1/x)}-4x ?
The limit as x approaches infinity of 4xe^{(1/x)}-4x is 4