What is the limit as x approaches infinity of (2^{-x}-2^x)/(2^{-x)+2^x} ?

The limit as x approaches infinity of (2^{-x}-2^x)/(2^{-x)+2^x} is -1

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limit as x approaches infinity of (2^{-x}-2^x)/(2^{-x)+2^x}

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

- 1

Solution steps

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

Divide by

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

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

x \to a lim [\frac{f ( x )}{g ( x )}] = \frac{lim _{x \to a} f ( x )}{lim _{x \to a} g ( x )}, x \to a lim g (x) \neq = 0

With the exception of indeterminate form

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

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

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

= \frac{- 1}{1}

Simplify

= - 1

\int \frac{2 x}{3 x ^{2} - 4} d x

\frac{3}{5 + x ^{2}}

n = 1 \sum \infty (\frac{- 2}{3})^{n}

\int 9 x ln (x) d x

x \to - 2 lim (4^{x} cos (π x))

What is the limit as x approaches infinity of (2^{-x}-2^x)/(2^{-x)+2^x} ?
The limit as x approaches infinity of (2^{-x}-2^x)/(2^{-x)+2^x} is -1