Final answer:
To find the value of a for which the limit exists, substitute the power series representation of sin(x) into the limit expression and simplify. The limit exists when a = 6 and evaluates to 0.
Step-by-step explanation:
To find the value of a for which the limit exists, we can use the power series representation of sin(x). The power series representation of sin(x) is x - (x^3)/6 + (x^5)/120 - .... We need to find the value of a that makes the limit lim (x → 0) [sin(ax) - sin(x) - 2x] / x^3 exist.
We can substitute the power series representation of sin(x) into the limit expression and simplify. After simplifying, we get the limit expression as lim (x → 0) [(ax^3)/6 + O(x^5)] / x^3. The term O(x^5) represents higher order terms that are negligible as x approaches 0.
Now, we can cancel out the x^3 term and find the limit as lim (x → 0) (ax)/6. The limit exists when a = 6. If a = 6, the limit evaluates to lim (x → 0) (6x)/6 = lim (x → 0) x = 0.