What is the limit as x approaches+0 of (sin(4x))/(tan(2x)) ?

The limit as x approaches+0 of (sin(4x))/(tan(2x)) is 2

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limit as x approaches+0 of (sin(4x))/(tan(2x))

x \to + 0 lim (\frac{sin ( 4 x )}{tan ( 2 x )})

2

Solution steps

x \to 0 lim (\frac{sin ( 4 x )}{tan ( 2 x )})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{cos ( 4 x ) \cdot 4}{sec ^{2} ( 2 x ) \cdot 2})

= x \to 0 lim (\frac{cos ( 4 x ) \cdot 4}{sec ^{2} ( 2 x ) \cdot 2})

Plug in the value

x = 0

= \frac{cos ( 4 \cdot 0 ) \cdot 4}{sec ^{2} ( 2 \cdot 0 ) \cdot 2}

Simplify

\frac{cos ( 4 \cdot 0 ) \cdot 4}{sec ^{2} ( 2 \cdot 0 ) \cdot 2} : 2

= 2

f (x) = a x + b

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

y = x (x^{4} + 1)

\int_{0}^{1} arcsin (w) d w

x \to 0 + lim (x^{x})

What is the limit as x approaches+0 of (sin(4x))/(tan(2x)) ?
The limit as x approaches+0 of (sin(4x))/(tan(2x)) is 2