# What is limit (1-cosx)^x if x go to 0?

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### 1 Answer

You need to evaluate the limit `lim_(x->0)(1 - cos x)^x` , hence, you need to replace 0 for x, such that:

`lim_(x->0)(1 - cos x)^x = (1 - cos 0)^0`

Since `cos 0 = 1` yields:

`lim_(x->0)(1 - cos x)^x = (1 - 1)^0`

`lim_(x->0)(1 - cos x)^x = 0^0`

The indetermination case `0^0` requests for you to use the following method, such that:

`lim_(x->0)(1 - cos x)^x = lim_(x->0) e^(ln((1 - cos x)^x))`

Using logarithmic identity `ln a^b = b*ln a` yields:

`lim_(x->0) e^(ln((1 - cos x)^x)) = e^(lim_(x->0) x*ln(1 - cos x))`

You should evaluate the limit such that:

`lim_(x->0) x*ln(1 - cos x) = 0*oo`

You need to convert the indetermination `0*oo` in `oo/oo` , such that:

`lim_(x->0) x*ln(1 - cos x) = lim_(x->0) (ln(1 - cos x))/(1/x) = oo/oo`

The indetermination case `oo/oo` requests for you to use l'Hospital's theorem, such that:

`lim_(x->0) (ln(1 - cos x))/(1/x) = lim_(x->0) ((ln(1 - cos x))')/((1/x)')`

`lim_(x->0) ((ln(1 - cos x))')/((1/x)') = lim_(x->0) (sin x/(1 - cos x))/((-1/x^2))`

Using the special limit `lim_(x->0) sin x/x = 1` yields:

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

The indetermination case `0/0` requests for you to use l'Hospital's theorem, such that:

`lim_(x->0) -x^3/(1 - cos x) = lim_(x->0) (-(x^3)')/((1 - cos x)')`

`lim_(x->0) ((-x^3)')/((1 - cos x)') = lim_(x->0) (-3x^2)/(sin x) = 0/0`

The indetermination case `0/0` requests for you to use l'Hospital's theorem again, such that:

`lim_(x->0) (-3x^2)/(sin x) = lim_(x->0) ((-3x^2)')/((sin x)') `

`lim_(x->0) ((-3x^2)')/((sin x)')= lim_(x->0) (-6x)/(cos x) = 0/1 = 0`

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

**Hence, evaluating the limit, under the given conditions, yields `lim_(x->0)(1 - cos x)^x = e^0 = 1` .**