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If `lim_(x->0) (a+bx sin x + c cosx)/x^4=2` then a=, b=, c=

Answer» `lim_(x->0) (a+bxsinx+c cosx)/x^4 = 2`
Here, we will use,
`sinx = x-x^3/(3!)+x^5/(5!)+...`
`cosx = 1-x^2/(2!)+x^4/(4!)+...`
So, given equation becomes,
`lim_(x->0) (a+(bx^2-(bx^4)/6+...)+(c-(cx^2)/2+(cx^2)/24+...))/x^4 = 2`
Now, terms of `x` with power less than `4` and more than `4` will be `0` .
`:. a+c = 0=> a = -c->(1)`
`b-c/2 = 0=>c = 2b->(2)`
Only terms with `x^4` will have some value.
` :. -b/6+c/24 = 2`
`=>-4b+c = 48`
From (2),
`=>-4b+2b = 48 => b = -24`
`=>c = 2**(-24) = -48`
`=a = -c = 48`


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