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

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

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

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

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{cos ( h + 2 x )}{1})

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

Plug in the value

h = 0

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

Simplify

= cos (2 x)

x \to 0 + lim (x^{\frac{6}{1 + 2 l n ( x )}})

x \to 0 - lim (3 e^{- \frac{9}{x}})

x \to 0 lim (1 x^{2})

x \to 4 + lim (2 + x - 4)

x \to 8 lim (7 - \frac{1}{2} x)

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