Sal finds the limit of (x²+x-6)/(x-2) at x=2 by factoring and simplifying the expression.. Created by Sal Khan.
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- Does l'Hopital's rule apply here?(77 votes)
- L'Hopitals rule is applicable here
L= lim x->2 for x^2+x-6/(x-2)
L= lim x->2 for f(x)/g(x) where f(x)=x^2+x-6, g(x)=x-2
since lim x->2 f(x)=0 and lim x->2 g(x)=0
and 0/0 is one of the inderminant forms we can apply L'Hopitals rule
L= lim x->2 for f'(x)/g'(x)=5/1=5
we obtained the same answer when we used factoring to solve the limit
In my opinion, it is easier to use L'Hopitals here than factoring (many will disagree with me).
However, you typically need to know limits before you learn calculus, and you need to know
calculus before you use L'Hopitals rule, so the 1st time people learn how to solve these
types of limits they will use factoring.(26 votes)
- How did Sal come up with a slope of 1?(36 votes)
- Consider y=mx+c where m is the slope of the line
Now see that when y=x+3 the value of m is 1 which means that slope is 1.(54 votes)
- One question, if you use algebraic rules on it, like sal did and you cancel the (x-2) out, after that the function f(x) is not the same as before or? Because then you dont have that gap anymore?(24 votes)
- Excellent question! And you're right, it's not exactly the same function, because the domain is different -- it now includes x=2, whereas the original function didn't. We can say that it's an extension of the original function, though, because it behaves exactly the same at all the points which were included in the original function.(42 votes)
- I'm still confuse, is it possible that calculus and algebra can be the same? Some problems i guess can be done through algebra like simple substitution of x=any number.(18 votes)
- Well, in the end, calculus is ultimately based on algebra. It's simply taking a few very useful algebraic functions, making a couple of modifications (like the addition of limits), and expanding on them until they've effectively become their own field of study, in the same way that algebra is ultimately based on arithmetic, but with the added concept of variables. You'd use arithmetic to solve an algebra problem, and you'd use algebra to solve a calculus problem (although, hopefully, you wouldn't use calculus to solve arithmetic). One of the most beautiful parts of mathematics is that it's all connected like that, with one step leading right to the next. So, while they are technically different areas of mathematics, in essence, yes, they are both a part the same thing. One is just a step above the other.(2 votes)
- Say I have lim of 5/x-7 as x -> 7 I know it is undefined but say I have lim 5/(x-7)^2 as x -> 7 why is it +infinity rather than undefined?(9 votes)
- When you're stuck like this, you need to see the limit when approaching to 7 for the right side and for the left. At the first example, lim x->7 (5/(x-7)), if you approach by the left you got -infinite, because you will have a number really close to 7, but still minor than it. And you approach by the right, you got + infinite, because you will have a number really close to 7, but a little higher than it. That's why it's undefined, because the two limits are different. If you do the same thing on the other one, you will get +infinite in the both sides, and the limit is defined as +infinite.(15 votes)
- I didn't get it...when we have f(x) = (x^2 + x -6)/x-2..then our limf(x)as x approaches 2 is not defined..but when w simplified the unction to x+3 then we got a different value of limit as x approaches 2 ..can we have two values for same limit..of the same function?(11 votes)
- the limit for the original equation is the same thing as the simplified version. we simply cant directly solve the limit in the original equation that easily so we simplify it first(6 votes)
lim f(x) = lim g(x)
x -> 1 x -> 1
can you say that
lim f(x)/g(x) = 1
when x -> 1
is always true?
or it would be correct ONLY as long as lim g(x) does not equal to 0 when x -> 1?
- You can't say that. Take for example f(x)=x-1 and g(x)=x^2 -1. Then
lim f(x) = lim g(x) = 0
x -> 1 x -> 1
However, lim f(x)/g(x) = lim (x-1)/(x^2 -1) = lim (x-1)/(x-1)(x+1) = lim 1/(x+1) = 1/2.
x -> 1 x -> 1 x -> 1 x -> 1(9 votes)
- If we factor x^2+x-6, f(x)=x+3. And why f(x) is still undefined at x=2?(2 votes)
- You can't just say that f(x) = x + 3 after canceling out (x - 2). That's still a part of the function, and if you just stay f(x) = x + 3 you lose out on the information that (x - 2) was in the denominator. That's still very much a part of the function, so (x+3)(x-2) / (x-2) is not equal to x + 3. It is equal to x + 3 everywhere but where x in the original function would make it undefined, and we need to show that as well by saying that f(x) = x + 3 everywhere but x = 2, where f(x) is undefined.(5 votes)
- Can someone give me some intuition as to why factoring works? I read somewhere that (x-2) is a "culprit" factor, which we need to factor out or something along those lines? Is this true, and if so, is there more to it than that?(2 votes)
- It is kind of because the function is undefined at x=2. But at a value of x infinitely close to 2(like 2.001 or1.999), we can evaluate the function. Hence we can calculate the limit of that function which is exactly that value which we can get very very close to but never exactly evaluate.(4 votes)
- Ok I looked over comments and there is one thing I do not understand. I get how to solve this for 5. I get how the graph starts with the X+3 and the 2 as the undefined location. I do not understand why he puts the circle at3:08where it is. I mean I understand how it will become 5 because that is the solution to this limit question. i do not understand why he places the circle there, he has not solved the equation yet, how does he know to place the circle there? He doesn't really say anything, just draws it there then says its 5 later on. Is there a trick or rule I am missing?(3 votes)
- The circle shows that the function is not defined when x=2. Sal knows this because the denominator is x-2 and anything divided by zero is undefined.(1 vote)