What is the limit as x approaches 0+of 1/x-cot(x) ?

The limit as x approaches 0+of 1/x-cot(x) is 0

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limit as x approaches 0+of 1/x-cot(x)

x \to 0 + lim (\frac{1}{x} - cot (x))

0

Solution steps

x \to 0 + lim (\frac{1}{x} - cot (x))

Multiply by the conjugate of

\frac{1}{x} - cot (x) : \frac{\frac{1}{x ^{2}} - cot ^{2} ( x )}{\frac{1}{x} + cot ( x )}

= x \to 0 + lim (\frac{\frac{1}{x ^{2}} - cot ^{2} ( x )}{\frac{1}{x} + cot ( x )})

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 _{x \to 0 +} ( \frac{1}{x ^{2}} - cot ^{2} ( x ) )}{lim _{x \to 0 +} ( \frac{1}{x} + cot ( x ) )}

x \to 0 + lim (\frac{1}{x ^{2}} - cot^{2} (x)) = \frac{2}{3}

x \to 0 + lim (\frac{1}{x} + cot (x)) = \infty

= \frac{\frac{2}{3}}{\infty}

Apply Infinity Property:

\frac{c}{\infty} = 0

= 0

\int \frac{5 f ( x )}{x ^{2} ( x + 1 ) ^{3}} d x

\int_{0}^{π} sin (n x) d x

\frac{- cos ( x )}{4 x ^{2} + 2 x - 3}

\frac{\partial}{\partial y} (ln (x + y^{3}))

\frac{d y}{d t} = 4 y

What is the limit as x approaches 0+of 1/x-cot(x) ?
The limit as x approaches 0+of 1/x-cot(x) is 0