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

The limit as x approaches-0+of (e^x+x)^{1/x} is e^2

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

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

e^{2}

Decimal

7.38905 \dots

Solution steps

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

Apply exponent rules

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

Apply the Limit Chain Rule:

e^{2}

= e^{2}

What is the limit as x approaches-0+of (e^x+x)^{1/x} ?
The limit as x approaches-0+of (e^x+x)^{1/x} is e^2