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

The limit as n approaches infinity of (n^2)/(2n-1)sin(1/n) is 1/2

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limit as n approaches infinity of (n^2)/(2n-1)sin(1/n)

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

\frac{1}{2}

Decimal

0.5

Solution steps

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

Divide by highest denominator power:

\frac{n}{2 - \frac{1}{n}}

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

Simplify

\frac{n}{2 - \frac{1}{n}} sin (\frac{1}{n}) : \frac{n ^{2} sin ( \frac{1}{n} )}{2 n - 1}

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

Divide by highest denominator power:

\frac{n sin ( \frac{1}{n} )}{2 - \frac{1}{n}}

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

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

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

= \frac{1}{2}

What is the limit as n approaches infinity of (n^2)/(2n-1)sin(1/n) ?
The limit as n approaches infinity of (n^2)/(2n-1)sin(1/n) is 1/2