What is the limit as n approaches infinity of 2^{n+1} ?

The limit as n approaches infinity of 2^{n+1} is infinity

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limit as n approaches infinity of 2^{n+1}

n \to \infty lim (2^{n + 1})

\infty

Solution steps

n \to \infty lim (2^{n + 1})

Apply exponent rules

= n \to \infty lim (e^{(n + 1) l n (2)})

Apply the Limit Chain Rule:

\infty

= \infty

d er i v a t i v e f (x) = \frac{5}{t ^{2}}

x \to 4 lim (\frac{13 x - 52}{2 x - 8})

\int \frac{x cos ( x ) - sin ( x )}{x ^{2}} d x

x \to 5 lim (3 x^{2} - 6 x + 8)

t an g e n t \frac{1}{x ^{2}} (1.1)

What is the limit as n approaches infinity of 2^{n+1} ?
The limit as n approaches infinity of 2^{n+1} is infinity