Approaching a Limit: Does x Reach 1?

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SUMMARY

The discussion centers on the concept of limits in calculus, specifically addressing whether a variable x approaching 1 actually reaches 1. The definition of a limit is clarified using the formal epsilon-delta definition. An example is provided with the function f(x) = (x^2 - 1)/(x - 1), demonstrating that the limit as x approaches 1 is 2, despite the function being undefined at x = 1. This illustrates that the limit's value is determined by the behavior of the function near the point of interest, not the value at that point.

PREREQUISITES
  • Understanding of calculus concepts, specifically limits
  • Familiarity with epsilon-delta definitions of limits
  • Basic knowledge of polynomial functions and their behaviors
  • Ability to manipulate algebraic expressions involving limits
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  • Study the epsilon-delta definition of limits in detail
  • Explore the concept of removable discontinuities in functions
  • Learn about L'Hôpital's Rule for evaluating indeterminate forms
  • Investigate the behavior of piecewise functions at their boundaries
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This discussion is beneficial for students of calculus, mathematics educators, and anyone seeking to deepen their understanding of limits and their applications in function analysis.

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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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Yes, you get as close as you can get.
 
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 <b>close</b> to 1 but <b>not equal</b> to 1, f(x) is close to 4.<br /> <br /> 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 <b>except</b> 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.[/itex][/itex][/itex]
 

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