What is the limit as x approaches infinity of \sqrt[x]{3^x+4^x} ?

The limit as x approaches infinity of \sqrt[x]{3^x+4^x} is 4

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limit as x approaches infinity of \sqrt[x]{3^x+4^x}

4

Solution steps

Apply exponent rules

= x \to \infty lim (e^{\frac{1}{x} l n (3^{x} + 4^{x})})

Simplify

e^{\frac{1}{x} l n (3^{x} + 4^{x})} : (3^{x} + 4^{x})^{\frac{1}{x}}

= x \to \infty lim ((3^{x} + 4^{x})^{\frac{1}{x}})

Apply exponent rules

= x \to \infty lim (e^{\frac{1}{x} l n (3^{x} + 4^{x})})

Apply the Limit Chain Rule:

4

= 4

\frac{d}{d x} (4 - 3 x^{2})

4 x^{\frac{5}{4}} + 2 x^{\frac{3}{2}} + 4 x

\int \frac{4 x ^{2} - 1}{( x ^{2} + 4 ) ( 2 - x )} d x

f (x) = - \frac{x}{4}

f (x) = 4 x^{2} - 4 x + 1, at x = 1, y = 0

What is the limit as x approaches infinity of \sqrt[x]{3^x+4^x} ?
The limit as x approaches infinity of \sqrt[x]{3^x+4^x} is 4