IDNLearn.com provides a seamless experience for finding and sharing answers. Ask any question and get a detailed, reliable answer from our community of experts.
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]
Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. Thank you for choosing IDNLearn.com. We’re committed to providing accurate answers, so visit us again soon.
