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visits member for 3 years, 3 months
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Jul
6
suggested rejected edit on Limits of $s(x)$ and $H(x)$
Jul
6
comment Mathematical Induction Problem with Fraction
@Kekker: So since $16 \neq 17$ does this equation hold for $n=2$? And what can you say about all $n$?
Jul
2
awarded  Curious
Jun
30
awarded  Popular Question
Jun
30
comment Prove that $\lim\limits_{x \to 0} \sinh(x)/x =1$.
I don't know if editing this would change the intent of this question... but, $\lim_{x \to 0} \frac{\sinh x}{x} \neq 0$
Jun
30
comment Prove that $\lim\limits_{x \to 0} \sinh(x)/x =1$.
\sin hx should be \sinh x... :)
Jun
29
comment Draw a function $g$ such that $g$ is defined on the interval $[-9,9]$
Are you sure the question said: $lim_{x\rightarrow 5} g(x)$ does not exist? Are you sure it isn't supposed to be $-5$?
Jun
28
comment Find $G'\left( x\right)$.
You don't want G^', you just need G'... That will stop the latex error.
Jun
28
comment How to interpret integration of a discontinous function
I believe it has to do with (informally speaking): Rationals are countably infinite, irrationals are uncountably infinite... Hence, the point discontinuities of the rationals are insignificant in the end. Which is why the integral is zero. Someone correct me if I'm wrong.
Jun
28
comment When are there no critical points?
@Gahawar: Wouldn't that function have a critical point everywhere? Since there exists no derivative?
Jun
28
revised Why is the function integrable?
added 2 characters in body
Jun
28
revised Why is the function integrable?
added 6 characters in body
Jun
28
answered Why is the function integrable?
Jun
27
comment What do I not understand about one-to-one functions?
@JBKing: Sorry I meant -1... Typed it too quickly.
Jun
27
accepted What do I not understand about one-to-one functions?
Jun
27
comment What do I not understand about one-to-one functions?
@JBKing: And thus, two-to-two, would mean that given a point, for any other arbitrary point it must not produce the same output! And in the case of x^2 it does, give x = 1, y = 1, let x_2 = -1, y = 1 thus it is not injective!
Jun
27
comment What do I not understand about one-to-one functions?
OHH! I think I get it know, it's misleading because most people just think of a normal function for every f(x) there is a unique y, but it should actually be that plus doing a "horizontal line" test? So, while x^2 is a function, it is not one-to-one, because it fails the horizontal line test. Whereas x^3 is?
Jun
27
comment What do I not understand about one-to-one functions?
I'm still a little bit confused, so if I am understanding, "thought incorrectly ... f(x) can only have one value", $f(x_1)$ can give multiple values, but $f(x_2)$ must give different values than $f(x_1)$ (given $x_1 \neq x_2$)? But one-to-one, does not imply that given one $x$, $f(x)$ gives one output?
Jun
27
comment What do I not understand about one-to-one functions?
Whoa, so an injective function is not necessarily a function, in terms of function vs relation? That is misleading. My high school terminology deceives me.
Jun
27
revised What do I not understand about one-to-one functions?
added 1 character in body