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Let $f(x):(a,b)\to R$ and $g(x):(a,b)\to R$ be continuous at the point $x_o$ $ϵ$ $(a,b)$. Then $f(x)/g(x)$ is continuous at the point $x_o$ $ϵ$ $(a,b)$ , for $g(x_0)$ different from zero. Proof: ...

That's not true in general.We can take a counterexample to show one case when it is not true: Let $f:A\to B$ and $g:B\to C$ be two surjective functions and let $h:A\to C$ be their composition such ...

Or we could just consider the function: $F(x)=f(x)-f(a)-\frac{f(b)-f(a)}{g(b)-g(a)}(g(x)-g(a))$ Clearly $F(x)$ is defined iff $g(b)$ is different from $g(a)$ , otherwise if $g(a)=g(b)$ then $F(x)$ ...

You know that $|x+y|\leq |x|+|y|$ holds for every real number , then taking $x=a-b$ and $y=b$ we get: $|a-b|\geq|a|-|b|$ ... (1) if we exchange $a$ and $b$ we'll have: $|a-b|\geq -(|a|-|b|)$, or $... View answer Accepted answer 1 votes a) In order to prove it using the axioms of multiplication first we use the axiom that states that there exists$1\in \mathbb R $such that for every$x\in \mathbb R $it holds that$x*1=x$,so we ... View answer 1 votes The speed of the cyclist is v1=16km/h and the speed of motorcyclist is v2=48km/h, when they two will meet the distance passed by each of them will be equal so d1=d2 but the time that took the ... View answer Accepted answer 1 votes Let x be an irrational number , then take two rational sequences an and bn (rational sequences means that every term of that sequence is a rational number) such that both of them converge to x and an ... View answer Accepted answer 0 votes From the definition of supremum we know that: 1.$A\geq a$for every$a\epsilon\{r \in \mathbb{R} | r>0, r^2<2\}$and 2.for every$\epsilon>0$there exists an$a\epsilon\{r \in ...

In mathematics, the qualifier pointwise is used to indicate that a certain property is defined by considering each value ${\displaystyle f(x)}$ of some function ${\displaystyle f.}$ An important ...

Starting from $2-1/n<2$ for every natural number $n$ we see that $2$ is an upper bound of the set $A={\{a_n|n∈N\}}$. We also see that for every $\epsilon>0$ there exists an $n ∈ N$ such that : ...

You can easily prove the first one using Archimedean Property for real numbers $\log m$ and $\log n$ And for the second one again use the Archimedean Property for real number $b*\log_{m} n$ Knowing ...

Let $\lim_{n\to\infty} (a_n)=a$ and $\lim_{n\to\infty} (b_n)$=b ,then using epsilon definition of limit and this property of real numbers: $|a-b|\leq|a|+|b|$ you can prove easily that $\lim_{n\to\... View answer 0 votes Based on an order axiom we know that if$ a$and$b $have the same sign then$a*b>0$let$a=b=1$, clearly a and be have the same sign therefore$a*b=1*1=1>0$View answer 0 votes You can see that if$T(n)=2\cdot T(n-1)+n$and$T(1)=1$then you'll have a sequence of$T(1),T(2),T(3),T(4),T(5),\dots$that looks like this:$1,4,11,26,57,120,\dots\$ and so on. From this you can see ...