- What does ∀x∃y actually mean? - Mathematics Stack Exchange
Here is the proof that statement 2 is true: Let P be some statement involving x and y "$\forall x,\forall y, P (x,y)$" means given any object x and any object y, the statement P is true about them Similarly, "$\forall y, \forall x, P (x,y)$" means given any object y, paired with any object x, the statement P is true about them Recall that if A and B are statements, the meaning of $$ A
- notation - What does ∈ mean? - Mathematics Stack Exchange
I have started seeing the "∈" symbol in math What exactly does it mean? I have tried googling it but google takes the symbol out of the search
- calculus - Finding $\int x^xdx$ - Mathematics Stack Exchange
These identities for $\int_0^1 x^ {-x}\ dx$ and $\int_0^1 x^x\ dx$ are sometimes called the "sophomore's dream" Look that up on Wikipedia
- Limit of $\frac {x^c-c^x} {x^x-c^c}$ as $x \rightarrow c$
My question is: Show that $\lim_ {x \rightarrow c} \frac {x^c-c^x} {x^x-c^c}$ exists and find its value Because the limit is 0 0 I've tried using L'Hopital's rule, but every time I differentiate it I still get 0 0?
- Proof of the derivative of $x^n$ - Mathematics Stack Exchange
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- Why the limit of $\\frac{\\sin(x)}{x}$ as $x$ approaches 0 is 1?
When you say x tends to $0$, you're already taking an approximation So, we have to calculate the limit here Taylor series gives very accurate approximation of sin (x), so it can be used to calculate limit
- Why is $\infty\times 0$ indeterminate? - Mathematics Stack Exchange
Any number, when multiplied by 0, gives 0 However, infinity is not a real number When we write something like $\infty \cdot 0$, this doesn't directly mean anything; rather, it's shorthand for a certain type of limit, where the first part approaches infinity Now, zero times anything approaching $\infty$ will still give a limit of zero However, that's not what the shorthand $\infty \cdot 0
- elementary number theory - Prove $ x^n-1= (x-1) (x^ {n-1}+x^ {n-2 . . .
My base case is when $n=2$ we have on the left side of the equation $x^2-1$ and on the right side: $ (x-1) (x+1)$ which when distributed is $x^2-1$ So my base case holds
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