evaluate; lim x tends to a (xsina-asinx) / (x-a)?

Answer 1

using l hospital rule , we get

sina-acosa

Answer 2

let l=lim(x->a) (xsina - asinx) / (x-a)
this is of form 0/0...............substituting x=a

therefore, we can apply L' Hospital's Rule; so differentiating numerator and denominator respectively.

=>l= lim(x->a) (sina - acosx)/1
=>l=sina - acosa
which is the required limit.

Answer 3

lim [ (xsin(a) - asin(x)) / (x - a) ]
x -> a

We have to use L'Hospital's rule to solve this. This means we take the derivative of each of the top and bottom and calculate a new limit. Keep in mind that "a" is a constant and can be differentiated accordingly as a constant.

lim [ (sin(a) - acos(x)) / (1) ]
x -> a

Now, we can safely plug in the value of a, giving us

sin(a) - a cos(a)

And this should be our answer.

Answer 4

Use L¨HÔPITAL
= limit sin(a) -acos(x) = sin(a) -a cos(a)