What is limit of (sinx)^(tgx) if x --->0?

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You need to replace 0 for x to evaluate the limit, such that:

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

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

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

Using power property of logarithms yields:

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

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

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

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

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

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

Replacing 0 for `lim_(x->0) tan x*ln sin x` yields:

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

Hence, evaluating the given limit, using l'Hospital's theorem, yields `lim_(x->0) (sin x)^tan x = 1` .

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