What is the limit as n approaches infinity of 1.04^n ?

The limit as n approaches infinity of 1.04^n is infinity

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limit as n approaches infinity of 1.04^n

n \to \infty lim (1.0 4^{n})

\infty

Solution steps

n \to \infty lim (1.0 4^{n})

Apply the common limit:

x \to \infty lim (a^{x}) = \infty, a > 1

= \infty

\frac{d}{d x} (- x^{3} + 6 x^{2} - 9 x)

\int \frac{- 5}{x ^{2} + 16} d x

\int \frac{2 e ^{- \frac{6}{x ^{4}}}}{x ^{5}} d x

\frac{d ^{2} y}{d t ^{2}} - 8 \frac{d y}{d t} + 16 y = 0

f (x) = x^{3}, at x = 1

What is the limit as n approaches infinity of 1.04^n ?
The limit as n approaches infinity of 1.04^n is infinity