True or False
Let a be a real number, and let f and g be real functions defined at all points x in some open interval containing a except possibly at x = a.
a) For each natural n, the function (x-a)^n[ sin(fx)*(x-a)^-n] has a limit as x → a.
False: Since x is not defined at a for at [sin(fx)*(x-a)^-n] since the denominator would be zero.
b) Suppose that {x_n} is a sequence converging to a with x_n not equal to a. If f(x_n) → L as n → ∞, then f(x) → L as x→ a.
True: By the sequential charaterization of limits, we know L = lim_(x → a) f(x) if and only if f(x_n) → L as n approaches infinity for every sequence x_n which converges to a.
c) IF f and g are finite valued on the open interval (a-1, a+1) and f(x) → 0 as x → a, then f(x)g(x) → 0 as x→ a.
I know the answer is true, but I don't understand why.
d) If lim_(x→a) f(x) does not exist and f(x) =< g(x) for all x in some open interval I contianing a, then lim_(x→a) g(x) doesn't exist neither.
False: Take f(x) = (x-1) which diverges, and g(x) = (x^2 -1)/(x-1) then limit of g(x) is 1.
Are these counterexamples acceptable? Please can anyone please help me verify this. I am preparing for an exam for today. Thank you.
In c) we have two functions which have a finite limit while $x \to a$. Thus, $\lim_{x \to a} f(x)g(x) = \lim_{x \to a} f(x) * \lim_{x \to a} g(x) = 0 * G = 0, \forall G \in \mathbb R$, where $G = lim_{x \to a}g(x)$.
What about d), I can say that limit of $\frac {1}{x-1}$ exists for $x = 1$ but is infinite, while a good example of function that really doesn't have any limit is $\sin\frac 1x, x \to 0$, however you can simply find a function (like $g(x) = 1$) majorating it which has a limit.