#education
for the love of the almighty God, someone explain this to me using the epsilon delta definition of limits. I have looked everywhere but i cant understand any of the explanations.
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Jul 26, 2024
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219tuktuk
727lim_x → 2 (2x + 5) = 2(2) + 5 = 4 + 5 = 9 x wede 2 betexega kuxer ye equestionu waga wede 9 nw eyetexega mihedew you can try by x=1 2x + 5»»»=7 x=1.3 y=7.6 x=1.6 y=8.2 x=1.9 y=8.8 x=1.95 y=8.9 x=1.99 y=8.98 as you can see as x approaches to 2 the equation f(x) or y approaches to 9
Jul 26, 2024LavenderCamel23
89Bro proof by epsilon delta method new yalew
Jul 26, 2024No_one_you_know
17Bro you don't need epsilon delta for this. First thing you have to do when doing limits is that check if it is continuous(not undefined) on the given number(which is 2 in this case). So since 2 is within in the domain of 2x+5(domain of all linear equation is all real numbers), you just have to put 2 in the equations which will give you: 2(2)+5=9 Which proves the problem above
Jul 26, 2024Tfdumean
238To explain why \(\lim_{{x \to 2}} (2x + 5) \neq 7\), let's evaluate the limit directly. The limit \(\lim_{{x \to 2}} (2x + 5)\) can be found by substituting \(x = 2\) into the expression \(2x + 5\): \[ \lim_{{x \to 2}} (2x + 5) = 2(2) + 5 = 4 + 5 = 9 \] Therefore: \[ \lim_{{x \to 2}} (2x + 5) = 9 \] Since \(9 \neq 7\), it is clear that \(\lim_{{x \to 2}} (2x + 5) \neq 7\).
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