Approaching a limit

  • Thread starter bacon
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Main Question or Discussion Point

When approaching a limit, say, x approaching 1, does x actually reach 1 or is it just infinitesimally close? In particular I'm interested in where the denominator of a function of interest contains the factor (x-1).

(It's hard to show a nice example without the Latex.)
 

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  • #2
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Yes, you get as close as you can get.
 
  • #3
HallsofIvy
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The definition of "limit" is "[itex]lim_{x\rightarrow a}= L[/itex] if and only if given any [itex]\epsilon> 0[/itex] there exist [itex]\delta> 0[/itex] such that if [itex]0< |x- a|< \delta[/itex] then [itex]|f(x)- f(a)|< \epsilon[/itex]. Notice the "0< |x-a|". What happens at x= a is irrelevant.

For example, if f(x)= x2+ 3 for all x except 1 and f(1)= 100000, then [itex]lim_{x\rightarrow 1} f(x)[itex] is still 3+ 1= 4: for any value of x close to 1 but not equal to 1, f(x) is close to 4.

Since you mention "In particular I'm interested in where the denominator of a function of interest contains the factor (x-1)", take [itex]f(x)= (x^2- 1)/(x-1). To find [itex]lim_{x\rightarrow 1}f(x)[/itex] note that [itex]x^2- 1= (x-1)(x+ 1)[/itex] so that [itex](x^2-1)/(x-1)= x+1[/itex] for all x except x= 1. Since the limit as x goes to 1 does not depend on the value at x= 1, the limit of [itex](x^2+ 1)/(x-1)[/itex] is the same as the limit of x+ 1 which, it is easy to see, is 1+ 1= 2.
 

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