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

The limit as x approaches infinity of 1+(2^{x+1})/(2^x-1) is 3

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

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

3

Solution steps

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

Simplify

2^{x + 1} : 2^{x} \cdot 2

= x \to \infty lim (1 + \frac{2 ^{x} \cdot 2}{2 ^{x} - 1})

x \to a lim [f (x) \pm g (x)] = x \to a lim f (x) \pm x \to a lim g (x)

With the exception of indeterminate form

= x \to \infty lim (1) + x \to \infty lim (\frac{2 ^{x} \cdot 2}{2 ^{x} - 1})

x \to \infty lim (1) = 1

x \to \infty lim (\frac{2 ^{x} \cdot 2}{2 ^{x} - 1}) = 2

= 1 + 2

Simplify

= 3

\int \frac{3}{y 4 - ( ln ( y ) ) ^{2}} d y

n = 1 \sum \infty \frac{1}{( n + 7 ) n}

f (x) = x - 2, (6, 2)

\frac{d y}{d x} 7 + e^{- 5 x} y = 0

f (x) = arctan (x^{2})

What is the limit as x approaches infinity of 1+(2^{x+1})/(2^x-1) ?
The limit as x approaches infinity of 1+(2^{x+1})/(2^x-1) is 3