# Approaching a limit

1. Apr 28, 2009

### bacon

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.)

2. Apr 28, 2009

### ImAnEngineer

Yes, you get as close as you can get.

3. Apr 28, 2009

### HallsofIvy

The definition of "limit" is "$lim_{x\rightarrow a}= L$ if and only if given any $\epsilon> 0$ there exist $\delta> 0$ such that if $0< |x- a|< \delta$ then $|f(x)- f(a)|< \epsilon$. 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 $lim_{x\rightarrow 1} f(x)[itex] is still 3+ 1= 4: for any value of x close to 1 but not equal to 1, f(x) is close to 4. Since you mention "In particular I'm interested in where the denominator of a function of interest contains the factor (x-1)", take [itex]f(x)= (x^2- 1)/(x-1). To find [itex]lim_{x\rightarrow 1}f(x)$ note that $x^2- 1= (x-1)(x+ 1)$ so that $(x^2-1)/(x-1)= x+1$ 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 $(x^2+ 1)/(x-1)$ is the same as the limit of x+ 1 which, it is easy to see, is 1+ 1= 2.