What is the limit as x approaches 0 of (e^{8x}-e^{-4x})/(ln(1+x)) ?

The limit as x approaches 0 of (e^{8x}-e^{-4x})/(ln(1+x)) is 12

Popular >

limit as x approaches 0 of (e^{8x}-e^{-4x})/(ln(1+x))

x \to 0 lim (\frac{e ^{8 x} - e ^{- 4 x}}{ln ( 1 + x )})

12

Solution steps

x \to 0 lim (\frac{e ^{8 x} - e ^{- 4 x}}{ln ( 1 + x )})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{8 e ^{8 x} + 4 e ^{- 4 x}}{\frac{1}{1 + x}})

= x \to 0 lim (\frac{8 e ^{8 x} + 4 e ^{- 4 x}}{\frac{1}{1 + x}})

Plug in the value

x = 0

= \frac{8 e ^{8 \cdot 0} + 4 e ^{- 4 \cdot 0}}{\frac{1}{1 + 0}}

Simplify

\frac{8 e ^{8 \cdot 0} + 4 e ^{- 4 \cdot 0}}{\frac{1}{1 + 0}} : 12

= 12

n \to \infty lim (\frac{2 n + 1}{1 - 3 n})

x \to 0 lim (\frac{ln ( 10 x + 1 )}{x})

x \to 0 lim (1 - \frac{tan ( 3 x )}{x})

x \to 4 + lim (10 - 2 x)

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

What is the limit as x approaches 0 of (e^{8x}-e^{-4x})/(ln(1+x)) ?
The limit as x approaches 0 of (e^{8x}-e^{-4x})/(ln(1+x)) is 12