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Sagot :
If [tex]f(x)[/tex] has a removable discontinuity at [tex]x=a[/tex], then the limit
[tex]\displaystyle \lim_{x\to a} \frac{f(x)}{x-a}[/tex]
exists and is finite.
A non-removable discontinuity at [tex]x=b[/tex] would entail a non-finite limit,
[tex]\displaystyle \lim_{x\to b} \frac{f(x)}{x-b} = \pm\infty[/tex]
or the limit does not exist (which could be due to the limits from either side of [tex]x=b[/tex] not matching or existing).
For a rational function, we want
[tex]f(x) = \dfrac{p(x)}{q(x)}[/tex]
where [tex]p[/tex] and [tex]q[/tex] are polynomials in [tex]x[/tex]. To get a removable discontinuity at [tex]x=a[/tex], both [tex]p[/tex] and [tex]q[/tex] must be divisible by [tex]x-a[/tex], and the limit of their quotient after removing these factors still exists. That is,
[tex]\displaystyle \lim_{x\to a} f(x) = \lim_{x\to a} \frac{p(x)}{q(x)} = \lim_{x\to a} \frac{(x-a)p^*(x)}{(x-a)q^*(x)} = \lim_{x\to a} \frac{p^*(x)}{q^*(x)} = \frac{p^*(a)}{q^*(a)}[/tex]
On the flip side, we get a non-removable discontinuity [tex]x=b[/tex] if [tex]p[/tex] is not divisible by [tex]x-b[/tex], in which case
[tex]\displaystyle \lim_{x\to b} f(x) = \lim_{x\to b} \frac{p(x)}{q(x)} = \lim_{x\to b} \frac{p(x)}{(x-b)q^*(x)} = \frac{p(b)}{0\times q^*(b)}[/tex]
and this is undefined.
Suppose [tex]f(x)[/tex] has a non-removable discontinuity at [tex]x=-5[/tex] and a removable one at [tex]x=4[/tex]. Then one such function could be
[tex]f(x) = \dfrac{x-4}{(x-4)(x+5)} = \dfrac{x-4}{x^2+x-20}[/tex]
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