What is the limit as x approaches 0+of 2^{1/x} ?

The limit as x approaches 0+of 2^{1/x} is infinity

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limit as x approaches 0+of 2^{1/x}

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

\infty

Solution steps

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

Apply exponent rules

= x \to 0 + lim (e^{\frac{1}{x} l n (2)})

Apply the Limit Chain Rule:

\infty

= \infty

\frac{2}{s ( 1 + 3 s )}

x = - 2, x = 1, y = 6 x, y = x^{2} - 7

n = 1 \sum \infty n^{5} (1 + n^{6})^{2}

\int e^{- x^{2} - 2 x + 7} (x + 1) d x

y^{^{''}} + 16 y = 3 sin (4 t), y (0) = 0, y^{^{'}} (0) = 0

What is the limit as x approaches 0+of 2^{1/x} ?
The limit as x approaches 0+of 2^{1/x} is infinity