What is the limit as x approaches infinity of (2^x+6^x)/(9^x-2^x) ?

The limit as x approaches infinity of (2^x+6^x)/(9^x-2^x) is 0

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limit as x approaches infinity of (2^x+6^x)/(9^x-2^x)

x \to \infty lim (\frac{2 ^{x} + 6 ^{x}}{9 ^{x} - 2 ^{x}})

0

Solution steps

x \to \infty lim (\frac{2 ^{x} + 6 ^{x}}{9 ^{x} - 2 ^{x}})

Divide by

9^{x} : \frac{( \frac{2}{9} ) ^{x} + ( \frac{2}{3} ) ^{x}}{1 - ( \frac{2}{9} ) ^{x}}

= x \to \infty lim (\frac{( \frac{2}{9} ) ^{x} + ( \frac{2}{3} ) ^{x}}{1 - ( \frac{2}{9} ) ^{x}})

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}{9} ) ^{x} + ( \frac{2}{3} ) ^{x} )}{lim _{x \to \infty} ( 1 - ( \frac{2}{9} ) ^{x} )}

x \to \infty lim ((\frac{2}{9})^{x} + (\frac{2}{3})^{x}) = 0

x \to \infty lim (1 - (\frac{2}{9})^{x}) = 1

= \frac{0}{1}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

x \to 2 lim (\frac{e ^{x} sin ( x )}{( x - 2 ) ^{3}})

x \to 0 lim (ln (1 - cos (x)))

x \to \infty lim ((1 + \frac{10}{x})^{5 x})

x \to 0 lim (\frac{1}{x} (\frac{1}{x + 4.25} - \frac{1}{4.25}))

x \to \infty lim (4 + 4 e \frac{1}{x - 3})

What is the limit as x approaches infinity of (2^x+6^x)/(9^x-2^x) ?
The limit as x approaches infinity of (2^x+6^x)/(9^x-2^x) is 0