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

The limit as h approaches 0 of ((sin(2(x+h))-sin(2x)))/h is 2cos(2x)

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limit as h approaches 0 of ((sin(2(x+h))-sin(2x)))/h

h \to 0 lim (\frac{( sin ( 2 ( x + h ) ) - sin ( 2 x ) )}{h})

2 cos (2 x)

Solution steps

h \to 0 lim (\frac{sin ( 2 ( x + h ) ) - sin ( 2 x )}{h})

Apply L'Hopital's Rule

: h \to 0 lim (\frac{2 cos ( 2 ( h + x ) )}{1})

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

Plug in the value

h = 0

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

Simplify

= 2 cos (2 x)

x \to π - lim (tan (\frac{x}{2}))

x \to π lim (x cos (x))

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

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

x \to 0 + lim (x - 2)

What is the limit as h approaches 0 of ((sin(2(x+h))-sin(2x)))/h ?
The limit as h approaches 0 of ((sin(2(x+h))-sin(2x)))/h is 2cos(2x)