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say we want the limit of x -> -4 for f(x) = (2x^2 + 6x + 8) / (x^2 + 4)
is it not possible to simplify the numerator? I am guessing I did the wrong thing when I simplified the numerator to 2((x + 2)(x + 4))?
So I have some questions.
1. Is it possible to simplify the numerator?
2. If the numerator was actually 2((x + 2)(x + 4))... would the limit of x -> -4 be 0.5?
3. Is the answer to the original question infinity?
Thanks guys.
2((x + 2)(x + 4)) = 2x² + 12x + 16, not 2x² + 6x + 8. The latter polynomial has no real roots, so it cannot be factored over reals.Quote:
Originally Posted by SplashDamage
No.Quote:
Originally Posted by SplashDamage
The limit would be 0 because x - 4 = 0 when x = -4.Quote:
Originally Posted by SplashDamage
No, it's 4/5. The denominator is never zero, so the function is continuous for all x, and its limit is its value.Quote:
Originally Posted by SplashDamage
Ok thanks. I see what I did wrong now. Also:
Basically I have no idea what this is trying to tell me. This is from the online lecture notes my lecturer put up. I have no idea what is going on, and no idea what h is.
http://desmond.imageshack.us/Himg407...ng&res=landing
1. Can someone please explain what is going on in steps, really simply.
2. what is h?
I know the basic idea is to make the two points closer and closer together, and then get the gradient of those two points when they are almost identical... But as you can see I don't get how to do it and what this is telling me.
There is not much to explain. P and Q are points on the parabola. The number h is the difference between the x-coordinates of P and Q. The slope of the line PQ, by definition, is the difference of the y-coordinates divided by the difference of the x-coordinates. Let's denote f(x) = x². It is assumed that P has coordinates (1, f(1)) = (1, 1), so Q has coordinates (1 + h, f(1 + h)) = (1 + h, (1 + h)²). Therefore, the slope is (f(1 + h) - f(1)) / h. As h tends to zero, the line PQ becomes the tangent to the parabola at P. (This last statement is informal at this point.)Quote:
Originally Posted by SplashDamage
Ah ok, so h is just the difference between x1 and x2?
Makes sense to me. Thanks man.
Since the function is defined and continuous for all x, you can evaluate this limit by substituting x = -4.Quote:
Originally Posted by SplashDamage
Here's a quick test: if x= -4, [itex]2x^2+ 6x+ 8= 2(-4)^2+ 6(-4)+ 8= 2(16)- 24+ 8= 32- 24+ 8= 8+ 8= 16. The fact that the value is not 0 tells you that x+ 4 is NOT a factor of the numerator. More to the point, since the denominator goes to 0 while the numerator does NOT shoulde tell you immediately that the limit does not exist.Quote:
Originally Posted by SplashDamage
IF the numerator had been [itex]2x^2+ 6x- 8[/itex], then with x= -4 we would have 2(16)- 24- 8= 8- 8= 0 and then we would know that x+ 4 is a factor: [itex]2x^2+ 6x- 8= 2(x^2+ 3x- 4)= 2(x+ 4)(x- 1)[/itex].
Quote:
So I have some questions.
1. Is it possible to simplify the numerator?
2. If the numerator was actually 2((x + 2)(x + 4))... would the limit of x -> -4 be 0.5?
3. Is the answer to the original question infinity?
Thanks guys.
The denominator does not go to zero, it's >= 4 for all x. If the denominator were x + 4 instead of x^2 + 4, then it would go to zero when x -> -4 .Quote:
Originally Posted by HallsofIvy
Quote:
Originally Posted by SplashDamage
as you may (or perhaps may not) recall, the slope of the line going through two points (x1,y1) and (x2,y2) is:
(y2 - y1)/(x2 - x1).
if our two points lie on the graph of a function f, such as f(x) = x2, then:
y2 = f(x2) = (x2)2
y1 = f(x1) = (x1)2.
suppose further, that we want to know what is happening "near x = a", so we take x1 = a, and let x2 = a+h (where h is how far away x2 is from x1).
so when x2 gets near x1, then a+h gets near a, that is: h must be getting near 0.
so now our slope formula becomes:
(y2 - y1)/(x2 - x1) = (f(x2) - f(x1))/(x2 - x1)
= (f(a+h) - f(a))/(a+h - a) = (f(a+h) - f(a))/h
now it looks like if we let h EQUAL 0, we have (f(a) - f(a))/0 = 0/0, which is bad, because 0/0 makes no sense.
but watch what happens if we just let h be NEAR 0, for f(x) = x2:
(f(a+h) - f(a))/h = ((a+h)2 - a2)/h = (a2 + 2ah + h2 - a2)/h
= (2ah + h2)/h = (2ah)/h + h2/h = 2a + h.
so when h is NEAR 0, the slope of our line is NEAR 2a.