What is the limit as x approaches 0 of ((4x^2))/(e^{4x)-4x-1} ?

The limit as x approaches 0 of ((4x^2))/(e^{4x)-4x-1} is 1/2

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limit as x approaches 0 of ((4x^2))/(e^{4x)-4x-1}

x \to 0 lim (\frac{( 4 x ^{2} )}{e ^{4 x} - 4 x - 1})

\frac{1}{2}

Decimal Notation

0.5

Solution steps

x \to 0 lim (\frac{4 x ^{2}}{e ^{4 x} - 4 x - 1})

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

= 4 \cdot x \to 0 lim (\frac{x ^{2}}{e ^{4 x} - 4 x - 1})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{2 x}{4 e ^{4 x} - 4})

= 4 \cdot x \to 0 lim (\frac{2 x}{4 e ^{4 x} - 4})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{2}{16 e ^{4 x}})

= 4 \cdot x \to 0 lim (\frac{2}{16 e ^{4 x}})

Plug in the value

x = 0

= 4 \cdot \frac{2}{16 e ^{4 \cdot 0}}

Simplify

4 \cdot \frac{2}{16 e ^{4 \cdot 0}} : \frac{1}{2}

= \frac{1}{2}

x \to 2 lim (\frac{3 x}{x ^{2} - 4})

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

x \to 0 + lim (1 + e^{- \frac{1}{x}})

x \to 0 lim (3 x + 2 + \frac{1}{x ^{2}})

x \to 0 lim (\frac{x - 1}{x + 2})

What is the limit as x approaches 0 of ((4x^2))/(e^{4x)-4x-1} ?
The limit as x approaches 0 of ((4x^2))/(e^{4x)-4x-1} is 1/2