- #1
fishingspree2
- 139
- 0
hello,
in my calculus introduction book, it is written:
Let a rational function: [tex]f(x) = \frac{{p(x)}}{{q(x)}}[/tex] and a, a real number
If q(a) equals 0, but not p(a), then [tex]{\lim }\limits_{x \to a} f(x)[/tex] does not exist.
however, while doing exercises on the internet, i found that:
[tex]{\lim }\limits_{x \to 1} \frac{{2 - x}}{{(x - 1)^2 }} = \infty[/tex]
is my textbook wrong?
thank you
in my calculus introduction book, it is written:
Let a rational function: [tex]f(x) = \frac{{p(x)}}{{q(x)}}[/tex] and a, a real number
If q(a) equals 0, but not p(a), then [tex]{\lim }\limits_{x \to a} f(x)[/tex] does not exist.
however, while doing exercises on the internet, i found that:
[tex]{\lim }\limits_{x \to 1} \frac{{2 - x}}{{(x - 1)^2 }} = \infty[/tex]
is my textbook wrong?
thank you