What is the limit as n approaches 0 of (sin(n))/n ?

The limit as n approaches 0 of (sin(n))/n is 1

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limit as n approaches 0 of (sin(n))/n

n \to 0 lim (\frac{sin ( n )}{n})

1

Solution steps

n \to 0 lim (\frac{sin ( n )}{n})

Apply L'Hopital's Rule

: n \to 0 lim (\frac{cos ( n )}{1})

= n \to 0 lim (\frac{cos ( n )}{1})

Plug in the value

n = 0

= \frac{cos ( 0 )}{1}

Simplify

\frac{cos ( 0 )}{1} : 1

= 1

\frac{d y}{d x} = \frac{y}{x} + \frac{y ^{2}}{x ^{2}}

\int (4 x + 3) cos (x) d x

y^{^{''}} + 3 y^{^{'}} + 2 = 0

\int_{1}^{9} e^{x} d x

k = 1 \sum \infty \frac{1}{k ^{4}}

What is the limit as n approaches 0 of (sin(n))/n ?
The limit as n approaches 0 of (sin(n))/n is 1