Find the values of a and b if f is continuous at x=0, where
`f(x)={:{((sinx + cos x)^(cosec x), , - pi/2 lt xlt 0),( a, , x=0 ), ((e^(1//x)+e(2//x)+e^(3//x))/(ae^(-2+1//x ) + be^(-1+3//x)),, 0 lt x lt pi/2):}`

Explanation

we have
f (0) =a
`lim_(x to 0-) f(x)= lim_(h to0) [sin (-h) + cos(-h)]^(cosec(-h))`
` = lim_(h to 0) [ cos h -sinh] ^(cosec h)`
`=lim_(h to 0) [1+ cosh - sinh -1]^(1/(cos h -sin h -1) xx (cos h - sin h -1)/(-sin h ))`
`Rightarrow lim_(x to0)f(x) = lim_(h to 0) [ 1+ cos h - sin h -1] ^((1/(cos h -sin h-1))xx((1+ sin h -cos h)/(sin h ))`
`lim_(h to 0) [1+(cos h -sin h -1) ^((1/(cos h - sin h -1))xxlim_(h to 0) ((2sin^(2)""h/2 + 2 sin ""h/2 cos""h/2)/(2 sin""h/2 cos""h/2)))`
`=e^(lim_(h to 0) (sin ""h/2 + cos ""h/2)/(cos""h/2)) =e " " [ because lim_(x to 0)(1+x)^(1/x) =e]`
Also `lim_(x to0+) f(x) = lim_( h to 0) (1/e^(h) + e^(2/h)+e^(3/h))/(ae^(-2+1/h)+be^(-1+3/h)) =lim_(h to 0) (e^((-2)/h + e^((-1)/h +1)+1))/ ((ae^(-2))e^((-2)/h) + (be^(-1)))`
`(0+0+1)/((ae^(-2))0+be^(-1)) = e/b`
For the function to be continuous at x = 0 we must have
`lim_(x to 0) f(x) =f(0) = lim_(x to 0+) f(x)`
` Rightarrow e =a = e/b`
` Rightarrow a=e and b =1`