What is the limit as x approaches 0 of (6^x+x)^{2/x} ?

The limit as x approaches 0 of (6^x+x)^{2/x} is e^{2(ln(6)+1)}

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limit as x approaches 0 of (6^x+x)^{2/x}

x \to 0 lim ((6^{x} + x)^{\frac{2}{x}})

e^{2 (l n (6) + 1)}

Decimal Notation

266.00601 \dots

Solution steps

x \to 0 lim ((6^{x} + x)^{\frac{2}{x}})

Apply exponent rules

= x \to 0 lim (e^{\frac{2}{x} l n (6^{x} + x)})

Apply the Limit Chain Rule:

e^{2 (l n (6) + 1)}

= e^{2 (l n (6) + 1)}

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

x \to - 3 lim (2 x^{2} - 3 x - 5)

x \to - \infty lim (x + 4)

x \to 0 lim (csch (x))

x \to 1 lim (\frac{2 x + 1}{x - 3})

What is the limit as x approaches 0 of (6^x+x)^{2/x} ?
The limit as x approaches 0 of (6^x+x)^{2/x} is e^{2(ln(6)+1)}