- #1
Juggler123
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I have to prove from the definiton of the limit of a function that the limit of f(x) as x tends to 2 equlas 3, given that;
f(x)=(2x-1)
I know that I have to find an epsilon such that |f(x)-l| [tex]\leq[/tex] [tex]\epsilon[/tex] and delta such that 0 [tex]\leq[/tex] |x-a| [tex]\leq[/tex] [tex]\delta[/tex]
Nowing putting in the conditions for this f(x);
|2x-4| [tex]\leq[/tex] [tex]\epsilon[/tex] and 0 [tex]\leq[/tex] |x-2| [tex]\leq[/tex] [tex]\delta[/tex]
But I don't where to go from here. Any help would be great!
f(x)=(2x-1)
I know that I have to find an epsilon such that |f(x)-l| [tex]\leq[/tex] [tex]\epsilon[/tex] and delta such that 0 [tex]\leq[/tex] |x-a| [tex]\leq[/tex] [tex]\delta[/tex]
Nowing putting in the conditions for this f(x);
|2x-4| [tex]\leq[/tex] [tex]\epsilon[/tex] and 0 [tex]\leq[/tex] |x-2| [tex]\leq[/tex] [tex]\delta[/tex]
But I don't where to go from here. Any help would be great!