What is the limit as x approaches infinity of (4^x+4^x)^{1/(2x)} ?

The limit as x approaches infinity of (4^x+4^x)^{1/(2x)} is 2

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limit as x approaches infinity of (4^x+4^x)^{1/(2x)}

x \to \infty lim ((4^{x} + 4^{x})^{\frac{1}{2 x}})

2

Solution steps

x \to \infty lim ((4^{x} + 4^{x})^{\frac{1}{2 x}})

Apply exponent rules

= x \to \infty lim (e^{\frac{1}{2 x} l n (4^{x} + 4^{x})})

Apply the Limit Chain Rule:

2

= 2

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x \to 0 lim (x^{6} cos (\frac{4}{x}))

\int_{0}^{\frac{π}{4}} 7 tan^{3} (x) d x

\int \frac{3 x}{( x ^{2} + 3 ) ^{2}} d x

f (x) = x + 2

What is the limit as x approaches infinity of (4^x+4^x)^{1/(2x)} ?
The limit as x approaches infinity of (4^x+4^x)^{1/(2x)} is 2