Question

Evaluate the limit as x approaches 0 of (1 - x^(sin(x)))/(x*log(x))

aka:
`(1-x^(sin(x)))/(x*log(x))`

Please include steps/explanation.

Answer 1

`sin~ x \approx x ~ ~\sf{as}~~ x \rightarrow 0`

We can replace sin x with x anywhere in the limit as long as x approaches 0.

Also,


`\large \lim_( x \to 0 ) ~ x^x = 1`

I will make the assumption that log(x)=ln(x).

The limit result can be proven if the base of log(x) is 10.


`\large \lim_(x \to 0^(+)) (1- x^(\sin x) )/(x \log x ) \\~\\ \large = \lim_(x \to 0^(+)) (1- x^(\sin x) )/( \log( x^x) ) \\~\\ \large = \lim_(x \to 0^(+)) (1- x^(x) )/( \log( x^x) ) ~~ \\ormalsize{\text{ substituting x for sin x } } \\~\\ \large = (\lim_(x \to 0^(+)) (1) - \lim_(x \to 0^(+)) \left( x^(x)\right) )/( \log( \lim_(x \to 0^(+))x^x) ) = (1-1)/(\log(1)) = (0)/(0)`

We get the indeterminate form 0/0, so we have to use Lhopitals rule


`\large \lim_(x \to 0^(+)) (1- x^(x) )/( \log( x^x) ) =_(LH) \lim_(x \to 0^(+)) (0 -x^x( 1 + \log (x)) )/(1 + \log (x) ) \\ = \large \lim_(x \to 0^(+)) (-x^x) = \large - \lim_(x \to 0^(+)) (x^x) = -1`

Therefore,


`\large \lim_(x \to 0^(+)) (1- x^(\sin x) )/(x \log x ) =\boxed{ -1}`