Calculate limit (e^sin2x) -1/3x with special limits (no hopital) if x go to 0?

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You should remember that `lim_(x->0) (e^(u(x)) - 1)/(u(x)) = 1` , hence, reasoning by analogy, `lim_(x->0) (e^(sin 2x) - 1)/(sin 2x) = 1` .

Since the given function is `(e^(sin 2x) - 1)/(3x), ` you need to multiply and divide by `sin 2x` such that:

`lim_(x->0) (e^(sin 2x) - 1)/(sin 2x) *(sin 2x)/(3x) = lim_(x->0) (e^(sin 2x) - 1)/(sin 2x) * lim_(x->0) (sin 2x)/(3x)`

You need to substitute 1 for `lim_(x->0) (e^(sin 2x) - 1)/(sin 2x)` such that:

`lim_(x->0) (e^(sin 2x) - 1)/(sin 2x) *(sin 2x)/(3x) = lim_(x->0) (sin 2x)/(3x)`

You need to remember the special limit `lim_(x->0) (sin 2x)/(2x) = 1` , hence, reasoning by analogy, yields:

`lim_(x->0) (sin 2x)/(2x) *(2x)/(3x) = 1*lim_(x->0) (2x)/(3x)=2/3`

`lim_(x->0) (e^(sin 2x) - 1)/(sin 2x) *(sin 2x)/(3x) =1*1*2/3`

`lim_(x->0) (e^(sin 2x) - 1)/(sin 2x) *(sin 2x)/(3x) = 2/3`

**Hence, evaluating the given limit, using the limits of special interest, yields `lim_(x->0) (e^(sin 2x) - 1)/(3x) = 2/3` .**

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