- #1
Oneiromancy
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I'm just kind of confused on how to handle the absolute value along with the inequality. Bad algebra skills. :(
0 < |x - x_0| < [tex]\delta[/tex] [tex]\Rightarrow[/tex] | f (x) - L | < [tex]\epsilon[/tex]
I'm given: f(x) = 2[tex]\sqrt{x + 1}[/tex], x[tex]_{0}[/tex] = 3, L = 4, [tex]\epsilon[/tex] = 0.2
The attempt at a solution
Nevermind, I tried latex and it messed everything up. Sorry.
0 < |x - x_0| < [tex]\delta[/tex] [tex]\Rightarrow[/tex] | f (x) - L | < [tex]\epsilon[/tex]
I'm given: f(x) = 2[tex]\sqrt{x + 1}[/tex], x[tex]_{0}[/tex] = 3, L = 4, [tex]\epsilon[/tex] = 0.2
The attempt at a solution
Nevermind, I tried latex and it messed everything up. Sorry.
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