Note,
(4-t^2)/(sqrt(4-t^2))=1/sqrt(4-t^2)
And when you approach t-->2 you have by substitution 1/0 which is an infinite limit.
Hello, ^_^Engineer_Adam^_^!
You can . . . but you get the same result.Why cant I use L'Hopital?
. . . . . . . . . . . . . . (4 - x²)^½
We have: . f(x) .= .--------------
. . . . . . . . . . . . . . . 4 - x²
. . . . . . . . . . . . .½(4 - x²)^{-½}·(-2x)
Apply L'Hopital: . ------------------------- . = . ½(4 - x²)^{-½}
. . . . . . . . . . . . . . . . . .-2x
. . . . . . . . . . . . . . . . . . . 1
Now take the limit of: .------------
. . . . . . . . . . . . . . . . .2√4 - x²
Technically, 1/0 is not infinity, it's undefined. In other words, 1/0 has no value.
However, if we use limits, there are some situations where we might get division by zero when plugging a specific value into a function, but we get infinity when x approaches that value. In other words,
If f(x) = 1/Q(x), where Q(a)=0
Then f(a) = 1/0 = undifined (no solution)
However,
If lim(x->a+)f(x) = infinity
And lim(x->a-)f(x) = infinity
Then lim(x->a)f(x) = infinity
In other words, 1/0 never equals infinity, but the idea that 1/0 'is the same as' infinity comes from the fact that the limits of funtions when they approach some point, which are undefined at that point, approach infinity.
I'm sorry if it sounds like I'm nit-picking; I'm not intending to. But the idea that "if a limit approaches 1/0 by definition it is an infinite limit" is not true. We are taught this in early calculus (and not really corrected in later calculus) because it's easy to remember, but as the value of the function approaches 1/0, the limit approaches infinity only under certain conditions: That the left and right hand limits of the function exist and are both equal to infinity. If any of these are not true, then 1/0 does not equal (approach) infinity.
for example:
lim(x->0) 1/x does not equal infinity because
lim(x->0+) 1/x = positive infinity
lim(x->0-) 1/x = negative infinity
There is no general limit of 1/x as x approaches 0. This is an example were as the function's value approaches 1/0, the limit does not exist (and for that reason is not equal infinity).
But I've gone WAY off topic, so just ignore everything I've said.