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

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

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

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

\frac{1}{3}

Decimal

0.33333 \dots

Solution steps

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

Divide by

3^{x + 1} : \frac{( \frac{2}{3} ) ^{x + 1} + \frac{1}{3}}{\frac{2 ^{x}}{3 ^{x + 1}} + 1}

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

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

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

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

Simplify

= \frac{1}{3}

x \to 0 + lim (- ln (x))

\int \frac{y ^{2} + 1}{y + 1} d y

\frac{d}{d x} (\frac{e ^{x}}{e ^{x} + 1})

x 8 - x^{2}

\int \frac{1}{( 4 + x ^{2} ) ^{\frac{3}{2}}} d x

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