What is the limit as b approaches 0 of cot(b)-csc(b) ?

The limit as b approaches 0 of cot(b)-csc(b) is 0

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limit as b approaches 0 of cot(b)-csc(b)

b \to 0 lim (cot (b) - csc (b))

0

Solution steps

b \to 0 lim (cot (b) - csc (b))

Multiply by the conjugate of

cot (b) - csc (b) : \frac{- 1}{cot ( b ) + csc ( b )}

= b \to 0 lim (\frac{- 1}{cot ( b ) + csc ( b )})

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

= - b \to 0 lim (\frac{1}{cot ( b ) + csc ( b )})

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 _{b \to 0} ( 1 )}{lim _{b \to 0} ( cot ( b ) + csc ( b ) )}

b \to 0 lim (1) = 1

b \to 0 lim (cot (b) + csc (b)) = \infty

= - \frac{1}{\infty}

Apply Infinity Property:

\frac{c}{\infty} = 0

= 0

x \to 7 - lim (\frac{x ^{2} - 4 x - 21}{∣ x - 7 ∣})

x \to 0 lim (\frac{x ^{3} - x}{x - 1})

x \to 2 lim (\frac{2 x - 4}{x + 3})

x \to 0 lim ([\frac{x}{a}] \cdot \frac{b}{x})

x \to 0 lim ([\frac{1}{x} - \frac{1}{ln ( 1 + x )}])

What is the limit as b approaches 0 of cot(b)-csc(b) ?
The limit as b approaches 0 of cot(b)-csc(b) is 0