What is limit of (ctg x)^sin x when x go to 0?

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You need to evaluate the limit to the function `f(x) = (cot x)^(sin x)` , hence, you need to replace 0 for x, such that:

`lim_(x->0) (cot x)^(sin x) = (cot 0)^sin 0 = 0^0`

ItTh is recommended in case of indetermination `0^0` to use the following special method, such that:

`lim_(x->0) (cot x)^(sin x) = lim_(x->0) e^(ln (cot x)^(sin x))`

Using the power property of logarithms, yields:

`lim_(x->0) e^(ln (cot x)^(sin x)) = lim_(x->0) e^(sin x*ln (cot x))`

`lim_(x->0) e^(sin x*ln (cot x)) = e^lim_(x->0)(sin x*ln (cot x))`

Evaluating the limit of exponent yields:

`lim_(x->0)(sin x*ln (cot x)) = sin 0*ln(cot 0) = 0*oo`

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

`lim_(x->0) (ln (cot x))/(1/sin x) = oo/oo`

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

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

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

`lim_(x->0) (1/cot x)/(cos x) = lim_(x->0) (tan x)/(cos x) `

`lim_(x->0) (tan x)/(cos x) = lim_(x->0) (sin x/cos x)/(cos x) `

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

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

**Hence, evaluating the limit, using the special method recommended in indetermination case `0^0` , yields `lim_(x->0) (cot x)^(sin x) = 1` .**

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