# Explain limit without hospitalslim (x goes to 0) (e^x-e^sinx)/(x-sinx)

*print*Print*list*Cite

### 1 Answer

You need to use the following remarkable limit such that:

`lim_(x->0) (e^(u(x)) - 1)/(u(x)) = 1`

You need to form the remarkable limit factoring out `e^(sin x)` such that:

`lim_(x->0) e^(sin x)(e^x/(e^(sin x)) - 1)/(x - sinx)`

You should use the following property of exponentials, such that:

`a^x/a^y = a^(x-y)`

Reasoning by analogy yields:

`lim_(x->0) e^(sin x)(e^(x - sin x) - 1)/(x - sinx)`

Notice that the exponent `x - sin x` coincides with denominator, hence, you may use the remarkable limit `lim_(x->0) (e^(u(x)) - 1)/(u(x)) = 1` such that:

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

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

Since `sin 0 = 0` yields:

`lim_(x->0) e^(sin x)(e^(x - sin x) - 1)/(x - sinx) = e^0`

`lim_(x->0) e^(sin x)(e^(x - sin x) - 1)/(x - sinx) = 1`

**Hence, evaluating the given limit, using the remarkable limit `lim_(x->0) (e^(u(x)) - 1)/(u(x)) = 1` , yields `lim_(x->0) (e^x - e^(sin x))/(x - sinx) = 1.` **