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

The limit as x approaches infinity of x/(3^{2x+1)} is 0

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

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

0

Solution steps

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

Apply L'Hopital's Rule

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

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

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

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

x \to \infty lim (1) = 1

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

= \frac{1}{2 ln ( 3 )} \cdot \frac{1}{\infty}

Simplify

\frac{1}{2 ln ( 3 )} \cdot \frac{1}{\infty} : 0

= 0

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What is the limit as x approaches infinity of x/(3^{2x+1)} ?
The limit as x approaches infinity of x/(3^{2x+1)} is 0