What is the limit as x approaches 0 of (tan(3x)cos(2x))/x ?

The limit as x approaches 0 of (tan(3x)cos(2x))/x is 3

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limit as x approaches 0 of (tan(3x)cos(2x))/x

x \to 0 lim (\frac{tan ( 3 x ) cos ( 2 x )}{x})

3

Solution steps

x \to 0 lim (\frac{tan ( 3 x ) cos ( 2 x )}{x})

\frac{tan ( 3 x ) cos ( 2 x )}{x} = tan (3 x) cos (2 x) \frac{1}{x}

= x \to 0 lim (tan (3 x) cos (2 x) \frac{1}{x})

Apply the following algebraic property

: a \cdot b = \frac{a}{\frac{1}{b}}, b \neq = 0

cos (2 x) \frac{1}{x} tan (3 x) = \frac{\frac{1}{x} tan ( 3 x )}{\frac{1}{c o s ( 2 x )}}

= x \to 0 lim (\frac{\frac{1}{x} tan ( 3 x )}{\frac{1}{c o s ( 2 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} tan ( 3 x ) )}{lim _{x \to 0} ( \frac{1}{c o s ( 2 x )} )}

x \to 0 lim (\frac{1}{x} tan (3 x)) = 3

x \to 0 lim (\frac{1}{cos ( 2 x )}) = 1

= \frac{3}{1}

Simplify

= 3

x \to \infty lim (16 x)

x \to 3 lim (6 x^{2} - 4 x + 1)

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

x \to 4 - lim (\frac{∣ 4 - x ∣}{4 - x})

x \to \infty lim (x - a)

What is the limit as x approaches 0 of (tan(3x)cos(2x))/x ?
The limit as x approaches 0 of (tan(3x)cos(2x))/x is 3