Proving a limit to infinity using epsilon-delta

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The limit of the function f(x) = 2x + 3 as x approaches infinity is proven to be infinity using the epsilon-delta definition of limits. Specifically, for any real number d, there exists a k such that if x > k, then f(x) > d. By selecting k = (d - 3)/2, the proof demonstrates that as x increases beyond this threshold, the function value exceeds any arbitrary d, confirming the limit is indeed infinity.

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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 $k=(d-3)/2$.
 

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