- #1
John O' Meara
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Ues the limit definition to prove that the stated limit is correct. [tex] \lim_{x->-2} \frac{1}{x+1}=-1[/tex]. The limit def' is |f(x)-L|<epsilon if 0<|x-a|< delta. So we have [tex]|\frac{1}{x+1} + 1| < \epsilon if 0 < |x- (-2)| < \delta \mbox{ therefore } |\frac{1}{x+1}||x+2| < \epslion \mbox{ if } 0 < |x + 2| < \delta[/tex]. To bound [tex] \frac{1}{x+1} \mbox { let } -1 < \delta < 1 [/tex].[tex] -1 < x+2< 1 \mbox{ that implies } -2 < x+1 < 0 \mbox{ therefore } \frac{-1}{2} > \frac{1}{x+1} > 0[/tex]. This cannot be correct as the answer for delta is [tex] \frac{\epsilon}{1+\epsilon}[/tex]. My algebra is rusty. Please help, thanks.