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

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

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

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

0

Solution steps

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

Apply exponent rules

= n \to \infty lim (e^{- n^{2} l n (1 + \frac{3}{n})})

Apply the Limit Chain Rule:

0

= 0

\int 4 x e^{4 x} d x

x \to 0 lim (\frac{( tan ( x ) )}{x})

\frac{d}{d t} (e^{2 t s i n (2 t)})

\frac{d}{d x} (\frac{x ^{2}}{8} - \frac{1}{x})

n = 5 \sum \infty n

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