MHB Proving a limit to infinity using epsilon-delta

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To prove that lim 2x + 3 = ∞ as x approaches ∞ using the epsilon-delta definition, one must show that for any real number d, there exists a k such that if x > k, then f(x) > d. The limit definition states that for every d, we can find a k such that x > k implies 2x + 3 > d. By rearranging the inequality, it is sufficient to choose k as (d - 3)/2. This approach confirms that the function grows without bound as x increases. Thus, the limit is proven using the epsilon-delta framework.
brooklysuse
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lim 2x + 3 = ∞.
x→∞

Pretty intuitive when considering the graph of the function. But how would I show this using the epsilon-delta definition?Thanks!
 
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Please expand the definition of the limit and write what you need to prove.
 
Looking to use this definition. f:A->R, A is a subset of R, (a, infinity) is a subset of A.

lim f(x) =infinity if for any d in R, there exists a k>a such that when x>k, then f(x)>d.
x->infinity
 
You may forget about $a$. So you have to prove that for every $d$ there exists a $k$ such that if $x>k$, then $f(x)=2x+3>d$. So consider an arbitrary $d$. You need to show that there exists a $k$ such that $x>k$ implies $x>(d-3)/2$. It's sufficient to take $k=(d-3)/2$.
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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