What is the limit as x approaches infinity of (3^x+2^x)/(3^{x+1)+2^{x-1}} ?

The limit as x approaches infinity of (3^x+2^x)/(3^{x+1)+2^{x-1}} is 1/3

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limit as x approaches infinity of (3^x+2^x)/(3^{x+1)+2^{x-1}}

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

\frac{1}{3}

Decimal Notation

0.33333 \dots

Solution steps

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

Divide by

3^{x + 1} : \frac{\frac{1}{3} + \frac{2 ^{x}}{3 ^{x + 1}}}{1 + \frac{2 ^{x - 1}}{3 ^{x + 1}}}

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

x \to a lim [\frac{f ( x )}{g ( x )}] = \frac{lim _{x \to a} f ( x )}{lim _{x \to a} g ( x )}, x \to a lim g (x) \neq = 0

With the exception of indeterminate form

= \frac{lim _{x \to \infty} ( \frac{1}{3} + \frac{2 ^{x}}{3 ^{x + 1}} )}{lim _{x \to \infty} ( 1 + \frac{2 ^{x - 1}}{3 ^{x + 1}} )}

x \to \infty lim (\frac{1}{3} + \frac{2 ^{x}}{3 ^{x + 1}}) = \frac{1}{3}

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

= \frac{\frac{1}{3}}{1}

Simplify

= \frac{1}{3}

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What is the limit as x approaches infinity of (3^x+2^x)/(3^{x+1)+2^{x-1}} ?
The limit as x approaches infinity of (3^x+2^x)/(3^{x+1)+2^{x-1}} is 1/3