What is the limit as x approaches 0+of 7(1/x-cot(x)) ?

The limit as x approaches 0+of 7(1/x-cot(x)) is 0

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limit as x approaches 0+of 7(1/x-cot(x))

x \to 0 + lim (7 (\frac{1}{x} - cot (x)))

0

Solution steps

x \to 0 + lim (7 (\frac{1}{x} - cot (x)))

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

= 7 \cdot 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 )}

= 7 \cdot 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

= 7 \cdot \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

= 7 \cdot \frac{\frac{2}{3}}{\infty}

Simplify

7 \cdot \frac{\frac{2}{3}}{\infty} : 0

= 0

y^{^{''}} + 9 y = sec^{2} (3 x)

y = x^{x}

x \to + π + lim (csc (x))

\frac{d}{d x} ((x - 42)^{2} - 3 x ln ((x - 100)^{2}))

(e^{x} y) d y = (e^{- y} + e^{- 2 x - y}) d x

What is the limit as x approaches 0+of 7(1/x-cot(x)) ?
The limit as x approaches 0+of 7(1/x-cot(x)) is 0