Final answer:
For part (a), it is true that lim x→0 f(x) = 0. We can use the definition of a limit to establish this claim. For part (b), the limit limx→0 (x*sinx(1/x)cos(1/x)) cannot be found using the limit laws, but it can be determined using the Squeeze Theorem.
Step-by-step explanation:
(a) Answer:
Given that for all n=1,2,..., f(1/10^n) = 0, we need to determine if it is true that lim x→0 f(x) = 0.
We can establish that the claim is true using the definition of a limit. According to the definition, for any ε > 0, there exists a δ > 0 such that if 0 < |x - 0| < δ, then |f(x) - 0| < ε.
In this case, since f(1/10^n) = 0 for all n=1,2,..., we can choose δ = 1/10^m, where m is any positive integer. Then, for any chosen ε > 0, if 0 < |x - 0| < δ = 1/10^m, we have |f(x) - 0| = |f(x)| = 0 < ε. Therefore, the claim is true and lim x→0 f(x) = 0.
(b) Answer:
We cannot use the limit laws to find limx→0 (xsinx(1/x)cos(1/x)) because the limit laws only apply to finite limits of individual functions. In this case, we have a product of multiple functions with limits involving 0.
To find this limit, we can use the Squeeze Theorem. If we can find two functions g(x) and h(x) such that g(x) ≤ x*sin(x*(1/x)*cos(1/x)) ≤ h(x) for all x in the domain, and both g(x) and h(x) have the same limit as x approaches 0, then the limit of the given expression will also be the same.
By simplifying the expression, we get g(x) = x*(-1) and h(x) = x*(1). Both g(x) = -x and h(x) = x have the limit of 0 as x approaches 0. Therefore, by the Squeeze Theorem, the limit limx→0 (x*sin(x*(1/x)*cos(1/x)) is also 0.