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

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

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

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

0

Solution steps

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

Divide by highest denominator power:

\frac{\frac{( - 1 ) ^{n + 1}}{n}}{2 - \frac{1}{n}}

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

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

n \to \infty lim (\frac{( - 1 ) ^{n + 1}}{n}) = 0

n \to \infty lim (2 - \frac{1}{n}) = 2

= \frac{0}{2}

Apply rule

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

= 0

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