What is the limit as x approaches-infinity of \sqrt[x]{2} ?

The limit as x approaches-infinity of \sqrt[x]{2} is 1

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limit as x approaches-infinity of \sqrt[x]{2}

1

Solution steps

Apply exponent rules

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

Simplify

e^{\frac{1}{x} l n (2)} : 2^{\frac{1}{x}}

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

Apply exponent rules

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

Apply the Limit Chain Rule:

1

= 1

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What is the limit as x approaches-infinity of \sqrt[x]{2} ?
The limit as x approaches-infinity of \sqrt[x]{2} is 1