duki said:
Ok, we're starting to look at infinite limits.
To be honest, I'm completely confused here...
One of the examples is y=\frac{x}{x^2-5x+4}
So we find the domain to be all reals where x is not 1 or 4... will this be the case for all equations like this? How do you know when to do it this way, and when to do it the other way, where a limit actually exists? How do you know when a limit doesn't exist?
The limit does not exists in two cases below:
The first case, is that the left-hand sided limit, and the right-hand sided limit are different. i.e the limit when x approaches some value
a from above (
x is close to a,
and x > a), and from below (
x is close to a,
and x < a) are
not the same.
You know this limit: \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 , \quad x \mbox{ in radians}, right? That means, No matter how x tends to 0 (above, or below), sin(x)/x will tend to 1.
The limit when x approaches 0, of the expression \frac{|\sin x|}{x} does not exist, because its two one-sided limits are different.
As x approaches 0 from above (when x is close to 0 enough from above, we can restrict x to be on the interval (0, pi/2), so that sin(x) is
positive), so we have:
\lim_{x \rightarrow 0 ^ +} \frac{|\sin x|}{x} = \lim_{x \rightarrow 0 ^ +} \frac{\sin x}{x} = 1
Whereas, as x approaches 0 from below (when x is close to 0 enough from below, we can restrict x to be on the interval (-pi/2, 0), so that sin(x) is
negative), that means:
\lim_{x \rightarrow 0 ^ -} \frac{|\sin x|}{x} = \lim_{x \rightarrow 0 ^ -} \frac{-\sin x}{x} = -1
So, the limit does not exist at x = 0.
Problem:
Ok, see if you can prove that the following limit DNE:
\lim_{x \rightarrow 0} \frac{|x|}{x}
The second case, is when the limit is infinity. You can either say that the limit is infinity, or the limit DNE (does not exist). Since infinity isn't actually a number.
The infinite limit occurs when the numerator tends to some number
not 0, and the denominator tends to 0.
In your example above, the limit does not exist at x = 1, and x = 4. Since, when x tends to 1, or 4, the denominator tends to 0, whereas the numerator tends to 1, and 4 respectively.
You can think like this: When you divide a number (
not 0), by some number that is close to 0, its value will get bigger, and bigger.
Say, 1/0.1 = 10, 1/0.01 = 100, 1/0.001 = 1000, ...
Note that, when the numerator, and denominator, both tend to 0, we have the Indeterminate Form 0/0. We must (rationalize, if it has square root, or cube root, ...), then factor, then do the cancellation, and, lastly, evaluate the limit.
Example:
\lim_{x \rightarrow 3} \frac{x - 3}{x ^ 2 - 4x + 3}
As x tends to 0, both numerator, and denominator tend to 0, so we must factor it.
\lim_{x \rightarrow 3} \frac{x - 3}{x ^ 2 - 4x + 3} = \lim_{x \rightarrow 3} \frac{x - 3}{(x - 3) (x - 1)} = \lim_{x \rightarrow 3} \frac{1}{x - 1} = \frac{1}{3 - 1} = \frac{1}{2}.
Then the part comes to find the limits, which require four equations. How do you determine the limits from the negative approach and the positive approach to be infinity? How are these equations different from the previous examples?
Thanks for you patience! :)
Can you show us the four equations you mentioned about?
You can have a quick look about
One-Sided Limit on wiki. It explains things quite well, there are even some graphs for visualization. Just skim through the page, and see if you can get it.
If there's something that troubles you, just post it here. :)