What is the limit as n approaches infinity of ((cos(1/n)-1))/(1/n) ?

The limit as n approaches infinity of ((cos(1/n)-1))/(1/n) is 0

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limit as n approaches infinity of ((cos(1/n)-1))/(1/n)

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

0

Solution steps

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

Simplify

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

: n \to \infty lim \frac{\frac{s i n ( \frac{1}{n} )}{n ^{2}}}{- \frac{1}{n ^{2}}}

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

Simplify

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

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

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

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

Apply the Limit Chain Rule:

0

= - 0

Simplify

= 0

\int - 5 t d t

\frac{d}{d x} (\frac{1 - 2 x}{3 + x})

\frac{d}{d x} (cos (x^{5}))

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

x \to 3 lim (x^{2} + 2 x)

What is the limit as n approaches infinity of ((cos(1/n)-1))/(1/n) ?
The limit as n approaches infinity of ((cos(1/n)-1))/(1/n) is 0