Notice that the function from denominator is `sin pi*x - ln x` and the function from numerator is `1 - x,` hence, if you substitute 1 for x to numerator and denominator yields:

`lim_(x-gt1) (1-x)/(sin pi*x - ln x) = (1-1)/(sin pi - ln 1) = 0/(0-0) = 0/0`

You should use the l'Hospital's theorem, hence you need to differentiate the numerator and denominator with respect to x, as independent functions, such that:

`lim_(x-gt1) ((1-x)')/((sin pi*x - ln x)') = lim_(x-gt1) (-1)/(pi*cos pi*x - 1/x)`

You need to substitute 1 for x such that:

`lim_(x-gt1) (-1)/(pi*cos pi*x - 1/x) = (-1)/(pi*cos pi - 1)`

`lim_(x-gt1) (-1)/(pi*cos pi*x - 1/x) = (-1)/(-pi-1)`

`lim_(x-gt1) (-1)/(pi*cos pi*x - 1/x) = 1/(pi+1) `

**Hence, evaluating the limit of the function `(1-x)/(sin pi*x - ln
x)` yields `lim_(x-gt1) (1-x)/(sin pi*x - ln x) = 1/(pi+1).`**

a)

`x^y = y^x`

using implicit differentiation,

`yx^(y-1) = xy^((x-1))*(dy)/(dx)`

`(dy)/(dx) = (yx^(y-1))/(xy^(x-1))`

`(dy)/(dx) = (x^(y-2))/(y^(x-2))`

## See eNotes Ad-Free

Start your **48-hour free trial** to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Already a member? Log in here.