is xsin(1/x) uniformly continuous on (0 1)?
436889
Note that the function xsin(1/x) is continuous as long as x == 0. Also, we have that |xsin(1/x)| ≤ |x|, so the squeeze theorem implies that limx→0 = 0.
745714
Since you cannot divide by zero, the result is undefined. So 1/0 is neither rational nor irrational. It is also neither real nor imaginary. This is the reason why 1/0 is undefined, it is simply not a number (note: some calculators show 1/0 as NaN which stands for Not a Number).
209936
At x = 0 the function is undefined, because there is a zero denominator. If x is positive then going closer and closer to zero keeps f(x) at 1. But if x is negative, going closer and closer to zero keeps f(x) at −1. So this function does not have a limit at x = 0.
1273470
(sinx)/x is undefined at x =0, so it is not analytic, and therefore it is a singular point. The power series cannot converge to an undefined value.
657455
The error is zero when the data fi are data points from a polynomial f(x) of degree less or equal to n because then the (n + 1)st derivative, f(n+1)(ξx), is identically zero. This is simply a statement of the uniqueness theorem of polynomial interpolation.
1126926
The function f is defined on the closed interval [0, 1] and satisfies f(0) = f(1). On the open interval (0, 1), f is continuous and strictly increasing. Which of the following statements is true? (A) f attains both a minimum value and a maximum value on the closed interval [0, 1].
356336
Well, If you are thinking Riemann integrable, Then every differentiable function is continuous and then integrable! However any bounded function with discontinuity in a single point is integrable but of course it is not differentiable!
1108262
Combining with existence thereom, the IVP (1) has unique solution y = y(x) defined in the interval |x − x0| ≤ α. |f(x, y1) − f(x, y2)| ≤ L|y1 − y2| ∀(x, yi) ∈ R. If ∂f/∂y exists and is bounded, then it necessarily saitisfies Lipschitz condition.
521644
The distribution function F is right continuous at some point a if and only if for every decreasing sequence of real numbers {xn}n≥1 such that xn↓a we have F(xn)↓F(a).
1456842
Indeed x/x is only defined when x≠0. And wherever it is defined, its value is 1.
1175847
As x → 0, xsin(1/x) → 0 but cos(1/x) oscillates between -1 and 1, so f (x) has no limit as x → 0 and f cannot be continuous at 0.
1008063
But actually in those points the function tan(x) is not defined and hence they are not in the domain of tan function. So, it is ok to say of course tan(x) is continuous in its domain.
255630
Yes, they are the same. You just have to factor the (x2 - 1) in the numerator of f(x) and cancel out the (x + 1).
1209331
Is everything we thought a lie? Not quite; if we look really close at x = -1, we see a hole in the graph, called a point of discontinuity. The line just skips over -1, so the line isn't continuous at that point. It's not as dramatic a discontinuity as a vertical asymptote, though.
560194
The graph of a cubic function has no jumps or holes. The function is continuous.
350951
When x = 0. Then the domain is "all x not equal to zero". The domain doesn't care what is in the numerator of a rational expression. The domain is only influenced by the zeroes of the denominator.
660162
Solution: (a) tan(x) is a continuous function on the closed interval [0, π/4], so by Theorem 19.2, it is uniformly continuous on [0, π/4].