You need to remember the fact that the limit of rational function is a constant term if the order of polynomial numerator and the order of polynomial denominator are alike.
Notice that the denominator is one term polynomial of first order, hence the numerator needs to have the same order such that: `f(x) = ax + b` .
You need to substitute ax + b for f(x) in limit such that:
`lim_(x-gt0) (ax+b)/x = 3`
Notice that substituting 0 for x in evaluation of limit yields `0/0` indeterminate case, hence you need to use l'Hospital rule such that:
`lim_(x-gt0) ((ax+b)')/(x') = lim_(x-gt0) a/1 = 3 =gt a = 3`
`` If the function `f(x) = (ax+b)*sin x =gt lim_(x-gt0) (ax+b)*(sin x)/x = 3 =gt lim_(x-gt0) (ax+b)*lim_(x-gt0) (sin x)/x = 3`
You need to remember that `lim_(x-gt0) (sin x)/x = 1` , hence `lim_(x-gt0) (ax+b) = 3.`
Substituting 0 for x in the limit yields `lim_(x-gt0) (ax+b) = a*0 + b = 3 =gt b = 3`
Hence, evaluating the function f(x) under the given conditions yields`f(x) = 3x + 3` .
`f'(x) = lim_(x-gt0) f(x)/x = (3x+3)' = 3`